It should be $n^k+mn^l+1$.

Rewrite as: \[ n^k + mn^\ell + 1 \mid n^{k+\ell} + n^k + mn^\ell. \]Now we split into cases (inevitable given the answer form): If $\ell \le k$, then we factor out $n^\ell$ to get \begin{align*} n^k + mn^\ell + 1 &\mid n^k + n^{k-\ell} + m \\ \implies n^k + mn^\ell + 1 &\mid n^{k-\ell} + m - mn^\ell - 1. \end{align*}In the last, since $\max(n^{k-\ell}+m, mn^\ell+1) < n^k+mn^\ell+1$, it follows the right-hand side must be zero. Consequently, $m = \frac{n^{k-\ell}-1}{n^\ell-1}$. In order for this to be an integer, we must have $\ell \mid k-\ell$, or $\ell \mid k$. Now suppose $\ell > k$. We get \[ n^k + mn^\ell + 1 \mid n^\ell + mn^{\ell-k} + 1. \]Now we divide into two sub-cases: If $m = 1$, then subtracting once more gives \[ n^k + mn^\ell + 1 \mid n^{\ell-k} - n^k. \]Since the right-hand side is clearly less than the left in absolute value, it follows $\ell-k = k$, or $\ell=2k$. If $m \ge 2$, then we contend that $n^k + mn^\ell + 1 > n^\ell + mn^{\ell-k} + 1 > 0$ which would be a contradiction. It would suffice to check $(m-1) n^\ell + n^k > m n^{\ell-k}$, but $n^\ell \ge 2n^{\ell-k}$ (since $n > 1$) and $(m-1) n^\ell > (m-1) n^{\ell-k}$, hence done. This completes all three cases, as desired.

Probably it is the easiest N4 in IMO Shortlist history.It was also asked in Azerbaijan TST .I solved it using some mod and inequality.(Similar to Evan Chen´s)

Unneccesarily long solution. We'll break the proof into several separate lemmas. Lemma 4.1 : If $N^{a} - 1 \mid N^{b} - 1$ and $a, b > 0$, then $a \mid b$. Proof : Suppose $b = qa + r$ where $0 \le r < a$. Then, $N^{a} - 1 \mid N^{qa + r} - 1$ and $N^{a} - 1 \mid N^{qa} - 1$ implies $N^{a} - 1 \mid N^{\gcd(qa, qa + r)} - 1 = N^{\gcd(qa, r)} - 1 \le N^{r} - 1$, which implies $r = 0$. Thus, $a \mid b$. Lemma 4.2 $L \le K \Rightarrow L \mid K, M = \frac{N^{K - L} - 1}{N^{L} - 1}$ Proof : $N^{K} + MN^{L} + 1 \mid N^{K + L} - 1$ and $N^{K} + MN^{L} + 1 \mid (N^{L} - 1)(N^{K} + MN^{L} + 1) = (N^{K + L} - 1) + MN^{2L} + N^{L} - N^{K} - MN^{L}$ implies $N^{K} + MN^{L} + 1 \mid N^{L}(MN^{L} + 1 - N^{K - L} - M)$. Since $(N^{K} + MN^{L} + 1, N) = 1$, we have $N^{K} + MN^{L} + 1 \mid MN^{L} + 1 - N^{K - L} - M$ $N^{K} + MN^{L} + 1 \mid MN^{L} + 1 - N^{K - L} - M \Rightarrow N^{K} + MN^{L} + 1 \mid MN^{L} + 1 - N^{K - L} - M - (N^{K} + MN^{L} + 1) = -(N^{K} + N^{K - L} + M)$. Thus, $N^{K} + MN^{L} + 1 \mid N^{K} + N^{K - L} + M$ However, RHS is positive, and also $2(N^{K} + MN^{L} + 1) = 2N^{K} + 2MN^{L} + 2 > N^{K} + N^{K - L} + M$. Thus, $N^{K} + MN^{L} + 1 = N^{K} + N^{K - L} + M \Rightarrow M = \frac{N^{K - L} - 1}{N^{L} - 1}$. By Lemma 4.1, $L \mid K - L \Rightarrow L \mid K$. This proves the claim. Lemma 4.3 $L = 2K \Rightarrow M = 1$ Proof : $(N^{K} + MN^{2K} + 1)(N^{K} - 1) = MN^{3K} - MN^{2K} + N^{2K} - 1$ Also, $N^{K} + MN^{2K} + 1 \mid M(N^{3K} - 1) = MN^{3K} - M$ Thus, $N^{K} + MN^{2K} + 1 \mid MN^{2K} - N^{2K} - M + 1 = (N^{2K} - 1)(M - 1)$ If $M > 1$, then $N^{K} + MN^{2K} + 1 > MN^{2K} - N^{2K} - M + 1 \iff N^{K} + M + N^{2K} > 0$, which is true. Thus, $M = 1$ must hold and this proves the claim. We've handled the cases $L \le K$ and $L = 2K$. For all the lemmas from now on, we assume $L \neq 2K$ and $L > K$. Lemma 4.4 : $N^{K} + MN^{L} + 1 \mid N^{2K} + N^{K} + M$ Proof : $(N^{K} + MN^{L} + 1)(N^{K} - 1) = N^{2K} + MN^{K + L} - MN^{L} - 1$. Since $N^{K} + MN^{L} + 1 \mid M(N^{K + L} - 1) = MN^{K + L} - M$, we have $N^{K} + MN^{L} + 1 \mid N^{2K} - MN^{L} - 1 + M$. Thus, $N^{K} + MN^{L} + 1 \mid (N^{2K} - MN^{L} - 1 + M) + (N^{K} + MN^{L} + 1) = N^{2K} + N^{K} + M$, as desired. Lemma 4.5 : $K < L < 2K$ Proof : We already assumed $L > K$ and $L \neq 2K$. Assume $L > 2K$. Then, $N^{K} + MN^{L} + 1 > N^{2K} + N^{K} + M \iff M(N^{L} - 1) > N^{2K} - 1$. However, $M(N^{L} - 1) \ge N^{2K + 1} - 1 > N^{2K} - 1$. Thus, the previous inequality is true. This contradicts Lemma 4.4. Thus, $L \le 2K$ which implies $K < L < 2K$. Lemma 4.6 : $N^{K} + MN^{L} + 1 \mid M(1 + MN^{L - K} + N^{L})$ Proof : $N^{K} + MN^{L} + 1 \mid M(N^{K + L} - 1) = MN^{K + L} - M$ and $N^{K} + MN^{L} + 1 \mid (N^{K} + MN^{L} + 1)(N^{K} - M) = N^{2K} - MN^{K} + MN^{K + L} - M^{2}N^{L} + N^{K} - M$, and thus we have $N^{K} + MN^{L} + 1 \mid N^{2K} - MN^{K} - M^{2}N^{L} + N^{K}$. But, $N^{2K} - MN^{K} - M^{2}N^{L} + N^{K} = N^{K}(N^{K} - M - M^{2}N^{L - K} + 1)$. Since $(N^{K} + MN^{L} + 1, N) = 1$, $N^{K} + MN^{L} + 1 \mid N^{K} - M - M^{2}N^{L - K} + 1 \Rightarrow N^{K} + MN^{L} + 1 \mid N^{K} - M - M^{2}N^{L - K} + 1 - (N^{K} + MN^{L} + 1) = -M(1 + MN^{L - K} + N^{L})$, and the conclusion follows. Lemma 4.7 : $M > N^{L - K}$ Proof : $0 \equiv M(N^{K} + MN^{L} + 1) \equiv M(N^{K} + MN^{L} + 1 - (1 + MN^{L - K} + N^{L})) \equiv M(N^{K} - N^{L} + MN^{L} - MN^{L - K}) \equiv MN^{L - K}(N^{2K - L} - N^{K} + MN^{K} - M) \pmod{(N^{K} + MN^{L} + 1)}$ (where Lemma 4.6 is used for the first congruence) Again, since $N$ is coprime with $N^{K} + MN^{L} + 1$, we have $N^{K} + MN^{L} + 1 \mid M(N^{2K - L} - N^{K} + MN^{K} - M)$. Next, we claim that $N^{2K - L} - N^{K} + MN^{K} - M > 0$. Indeed, $N^{2K - L} - N^{K} + MN^{K} - M > 0 \iff M(N^{K} - 1) > N^{K} - N^{2K - L}$. But $M(N^{K} - 1) \ge N^{K} - 1 > N^{K} - N^{2K - L}$, so the claim is true. Now, we must have $M(N^{2K - L} - N^{K} + MN^{K} - M) \ge N^{K} + MN^{L} + 1$, or equivalently $M \ge \frac{N^{K} + MN^{L} + 1}{N^{2K - L} - N^{K} + MN^{K} - M}$. However, $N^{K} + MN^{L} + 1 > N^{L - K}(MN^{K} - M - N^{K} + N^{2K - L}) \iff N^{K} + MN^{L} + 1 > MN^{L} - MN^{L - K} - N^{L} + N^{K} \iff MN^{L - K} + N^{L} + 1 > 0$, which is true. Thus, $M \ge \frac{N^{K} + MN^{L} + 1}{N^{2K - L} - N^{K} + MN^{K} - M} > N^{L - K}$, as desired. Lemma 4.8 : $M \le N^{2K - L}$ Proof : By Lemma 4.4, we have $N^{2K} + N^{K} + M \ge N^{K} + MN^{L} + 1 \Rightarrow M(N^{L} - 1) \le N^{2K} - 1 \Rightarrow M \le \frac{N^{2K} - 1}{N^{L} - 1}$. However, $(N^{2K} - 1) < (N^{L} - 1)(N^{2K - L} + 1) \iff N^{2K} - 1 < N^{2K} + N^{L} - N^{2K - L} - 1 \iff N^{L} > N^{2K - L}$, which is true as $L > 2K - L$. Thus, $M \le \frac{N^{2K} - 1}{N^{L} - 1} < N^{2K - L} + 1$. Since $M$ is an integer, $M \le N^{2K - L}$, as desired. Lemma 4.9 : $M \mid N^{K} + 1$ Proof : Let $g = \gcd(N^{K} + 1, M) = \gcd(N^{K} + MN^{L} + 1, M)$. Then, by Lemma 4.6, $\frac{N^{K} + MN^{L} + 1}{g} \mid \frac{M}{g}(1 + MN^{L - K} + N^{L})$ $\Rightarrow \frac{N^{K} + MN^{L} + 1}{g} \mid 1 + MN^{L - K} + N^{L}$ (as $\gcd(\frac{N^{K} + MN^{L} + 1}{g}, \frac{M}{g}) = 1$) $\Rightarrow N^{K} + MN^{L} + 1 \mid g(1 + MN^{L - K} + N^{L})$ This implies $g(1 + MN^{L - K} + N^{L}) \ge N^{K} + MN^{L} + 1$. Note that $MN^{L - K} + N^{L} + 1 \le N^{2K - L} \cdot N^{L - K} + N^{L} + 1 = N^{K} + N^{L} + 1 < 2N^{L}$. (We used Lemma 4.8 for the first inequality) Thus, $g \ge \frac{N^{K} + MN^{L} + 1}{1 + MN^{L - K} + N^{L}} > \frac{N^{K} + MN^{L} + 1}{2N^{L}} > \frac{MN^{L}}{2N^{L}} = \frac{M}{2}$. However, recall that $g \mid M$ by definition, and thus $g > \frac{M}{2}$ implies $g = M$ (the second largest divisor of $M$ must be at most $\frac{M}{2})$. This implies $M \mid N^{K} + 1$, as desired. \end{proof} Armed with all these lemmas, we're ready to complete the solution. Let $G = \frac{N^{K} + 1}{M}$. $N^{K} + MN^{L} + 1 \mid N^{K + L} - 1 \Rightarrow GM + MN^{L} \mid N^{K + L} - 1 \Rightarrow M(G + N^{L}) \mid N^{K + L} - 1$. In particular, this implies $M \mid N^{K + L} - 1$. However, by Lemma 4.9, we have $M \mid N^{K} + 1 \mid N^{2K} - 1$, so $M \mid N^{\gcd(2K, K + L)} - 1 = N^{\gcd(2K, L - K)} - 1 \mid N^{L - K} - 1$. However, by Lemma 4.7, we have $M > N^{L - K}$, and $N^{L - K} - 1 > 0$, so we have a contradiction. In conclusion, the only possible solutions are $L \mid K, M = \frac{N^{K - L} - 1}{N^{L} - 1}$ and $L = 2K, M = 1$ from Lemma 4.2 and Lemma 4.3.

Relabel $n,m,k,l$ as $a,b,c,d$. Consider two cases. \underline {Case I: $c\ge d$} $$a^c+ba^d+1|a^{c+d}-1$$$$a^c+ba^d+1|a^{c+d}+a^c+ba^d$$$$\gcd(a^c+ba^d+1,a)=1 \Rightarrow a^c+ba^d+1|a^c+a^{c-d}+b$$$$a^c+ba^d+1\le a^c+a^{c-d}+b \Rightarrow b\le \displaystyle\frac{a^{c-d}-1}{a^d-1}$$If $b<\displaystyle\frac{a^{c-d}-1}{a^d-1}$, then $$a^c+a^{c-d}+b>a^c+ba^d+1$$$$a^c+a^{c-d}+b\ge 2(a^c+ba^d+1)$$$$a^{c-d}+b\ge a^c+2ba^d+2$$False since $a^{c-d}\le a^c, b<b+2<2ba^d+2$. So $b=\displaystyle \frac{a^{c-d}-1}{a^d-1}\Rightarrow a^d-1|a^{c-d}-1 \Rightarrow d|c-d \Rightarrow d|c$. This is a solution given in (ii). \underline {Case II: $c< d$} $$a^c+ba^d+1|a^{c+d}-1$$$$a^c+ba^d+1|ba^{c+d}-b-(a^{2c}+ba^{c+d}+a^c)$$$$a^c+ba^d+1|a^{2c}+a^c+b$$$$a^c+ba^d+1\le a^{2c}+a^c+b \Rightarrow a^d-1\le b(a^d-1)\le a^{2c}-1 \Rightarrow d\le 2c$$On the other hand, $$a^c+ba^d+1|a^{c+d}+a^c+ba^d$$$$\gcd(a^c+ba^d+1,a)=1 \Rightarrow a^c+ba^d+1|a^d+1+ba^{d-c}$$$$a^c+ba^d+1\le a^d+1+ba^{d-c}$$$$a^d-a^{d-c}\le b(a^d-a^{d-c})\le a^d-a^c \Rightarrow a^c\le a^{d-c} \Rightarrow 2c\le d $$So $d=2c$, then since $b(a^{2c}-1)=b(a^d-1)\le a^{2c}-1$, then $b\le1$. So $b=1$ since $b$ is a positive integer and $d=2c$. This is a solution given in (i). We conclude that there is no other solutions other than (i) and (ii). Q.E.D

Follows from the following generalized result (apply the case $(p,q,s,t) = (1,m,k,\ell)$ if $k\geq\ell$, and the case $(p,q,s,t) = (m,1,\ell,k)$ if $k\leq\ell$): Proposition. If $n$, $p$, $q$, $s$, $t$ are positive integers, with $n > 1$ and $s\geq t$, such that $pn^s + qn^t+1$ divides $n^{s+t}-1$, then $t | s$, $p = 1$ and $q = \frac{n^{s-t}-1}{n^t-1}$. Proof. Let $a = \frac{n^{s+t}-1}{pn^s+qn^t+1}$ be the quotient. Note that $$ a = \frac{n^{s+t}-1}{pn^s+qn^t+1} < \frac{n^{s+t}}{pn^s} = \frac{n^t}{p}\hspace{4mm}\text{and}\hspace{4mm}a = \frac{n^{s+t}-1}{pn^s+qn^t+1}\equiv \frac{-1}{1} = -1\pmod{n^t}, $$so $p = 1$ and $a = n^t-1$, and substituting these back into the definition of $a$ we get $q = \frac{n^{s-t}-1}{n^t-1}$. That $q$ is an integer implies $t|(s-t)$, so $t|s$.

This was proposed by Mexico.

This problem is too easy for IMO SL N4 ,in my opinion.

$\underline{solution}$ We have two cases $\boxed{Case-1}$: $\boxed{l>k}$ If we have $m=1$ $\implies n^k+n^l+1|n^{k+l}-1$ $\implies n^k+n^l+1|n^{k+l}-1-n^{2k}-n^{k+l}-n^k$ $\implies n^k+n^l+1|n^{2k}+n^k+1$ $\implies n^k+n^l+1|n^{2k}-n^l$ Now we have two sub cases $\boxed{l<2k}$ In this case,$n^k+n^l+1|n^{2k-l}-1$ $\implies n^{2k-l}-1\geq n^k+n^l+1$ $$n^{2k-l}>n^k+n^l$$$$n^{k-l}>n^{l-k}$$$$2k>2l$$a contradiction. $\boxed{l>2k}$ $\implies n^k+n^l+1|n^{l-2k}-1$ $\implies n^{l-2k}-1\geq n^k+n^l+1$ $\implies n^{l-2k}>n^k+n^l$ $\implies n^{l-3k}>n^{l-k}$ A contradiction. So we conclude for $m=1$,$l=2k$ Now we need to check for other m $n^k+mn^l+1|n^{k+l}-1$ $\implies n^k+mn^l+1|n^{k+l}+mn^l+n^k-(1)$ Now as$(n^k+mn^l+1,n^k)=1$ $n^k+mn^l+1|n^l+mn^{l-k}+1$ Some computation gives $m\leq\frac{n^l-n^k}{n^l-n^{l-k}}$ $\implies \frac{n^l-n^k}{n^l-n^{l-k}}\geq m\geq 2$ $\implies n^l-n^k\geq 2n^l-2n^{l-k}$ $\implies 2n^{l-k}\geq n^l+n^k$ $\implies 2n^{l-2k}>n^{l-k}$ $\implies 2>n^k$ which is absurd.$\square$ $\boxed{Case-2}$ $\boxed{l\leq k}$ $n^k+mn^l+1|n^{k+l}-1$ $n^k+mn^l+1|n^{k+l}-1-(n^l(n^k+mn^l+1))$ $n^k+mn^l+1|mn^{2l}+n^l+1$ $\implies mn^{2l}+n^l\geq n^k+mn^l$ Taking n^l as common $\implies m\geq\frac{n^{k-l}-1}{n^l-1}-(2)$ Now on the other hand $n^k+mn^l+1|n^{k+l}+mn^l+n^k$ from$(1)$ $n^k+mn^l+1|n^k+m+n^{k-l}$ $\implies n^k+m+n^{k-l}\geq n^k+mn^l+1$ $\implies m+n^{k-l}\geq mn^l+1$ $\implies \frac{n^{k-l}-1}{n^l-1}\geq m-(3)$ Thus we conclude from$(2)$ and $(3)$, that $m=\frac{n^{k-l}-1}{n^l-1}$ $\implies n^l-1|n^{k-l}-1$,$l|k-l$,$l|k$$\square$ Hence $Q.E.D$

oops, my solution is identical to that of v_Enhance. Posting anyway for storage. Let $M=n^k+mn^\ell+1$. Working modulo $M$, we have that \[n^{k+\ell}+n^k+mn^\ell\equiv 0\pmod{M}.\]Since $\gcd(n,M)=1$, we can divide out the highest power of $n$ dividing the left side. To do this, we need to split into two cases. Case 1: $\ell\le k$ Here, we have that \[n^k+n^{k-\ell}+m\equiv 0\pmod{M}.\]Suppose the left side was not equal to $M$. Then we have that the left side is at least $2M$, so \[n^k+n^{k-\ell}+m\ge 2(n^k+mn^\ell+1) \implies n^{k-\ell}+m\ge n^k+2mn^\ell+2.\]But $n^{k-\ell}\le n^k$ and $m\le 2mn^\ell$, so we have a contradiction. Therefore, we have \[n^k+n^{k-\ell}+m=M\implies n^{k-\ell}-1=m(n^\ell-1)\implies m=\frac{n^{k-\ell}-1}{n^\ell-1}.\]For this to be an integer, we must have $\ell\mid k-\ell$, or $\ell\mid k$. This takes care of the first solution possibility. Case 2: $\ell>k$. Here, we get that \[n^\ell+1+mn^{\ell-k}\equiv 0\pmod{M},\]or \[1+n^k+mn^\ell\mid 1+n^\ell+mn^{\ell-k}.\]In particular, the left side is at most the right, so we have \[n^k+mn^\ell\le n^\ell+mn^{\ell-k},\]or \[1-\frac{1}{n^{\ell-k}}\ge m\left(1-\frac{1}{n^k}\right).\]We see that $1/n^k\le 1/2$ since $n\ge 2$, so the left side is at least $m/2$. This is a clear contradiction for $m\ge 2$, so we must have $m=1$. We then get $\ell-k=k$, so $\ell=2k$. This completes the other solution possibility. $\blacksquare$

$n^k + mn^l+1 \mid n^{k+l}-1 \Longleftrightarrow n^k+mn^l+1 \mid n^{k+l}+n^k +mn^l $ (*) Note that $\gcd (n^k +mn^l +1, n^{k+l}+n^k+mn^l)=1$. We have two cases. Case 1: $k \geq l$ $n^k+mn^l+1 \mid n^k+n^{k-l}+m $ but $n^k +n^{k-l}+m \leq n^k +mn^l +1$. $n^k +mn^l +1 = n^{k-l}+m \Longrightarrow m = \frac{n^{k-l}-1}{n^l-1}$. Since $m$ is an integer $n^l-1 \mid n^{k-l}-1$ which implies $l \mid l- k \Longrightarrow l \mid k$. Case 2: $l>k$ Let $l=ak+r , r<k$ We have two subcases, Subcase 2.1: $a \geq 2$ From (*) $$n^l +mn^{l-k}+1 \equiv n^l +mn^{l-k} -n^k - mn^l \equiv n^{l-k} +mn^{l-2k} -mn^{l-k} -1$$$$\equiv n^{l-k}+mn^{l-2k}-mn^{l-k}+mn^l + n^k \equiv n^{l-2k} + mn^{l-3k} -mn^{l-2k} + mn^{l-k} +1 $$$$\equiv \dots \equiv n^{l-(a-1)k} +mn^{l-ak} - mn^{l-(a-1)k} +mn^{l-(a-2)k} + \dots + (-1)^{a} mn^{l-k} +(-1)^{a+ 1} \mod(n^k +mn^l +1)$$But, $$n^{l-(a-1)k} +mn^{l-ak} - mn^{l-(a-1)k} +mn^{l-(a-2)k} + \dots + (-1)^{a} mn^{l-k} +(-1)^{a+ 1} < n^{l-ak} + mn^{l-ak} = n^{k+r} + mn^{r} < n^{2k}+mn^r +1$$We will prove the following inequality $$n^{2k} +mn^r+1 < n^k +mn^l +1 \Longleftrightarrow n^{2k}-n^k < mn^{ak+r} -mn^r \Longleftrightarrow n^k(n^k-1)< mn^{ak}-m \Longleftrightarrow n^k< m(n^{(a-1)k} + n^{(a-2)k} + \dots + 1) $$Clearly $$m(n^{(a-1)k} + n^{(a-2)k} + \dots + 1) > mn^k +m > n^k$$So the inequality holds. Hence, $ n^{l-(a-1)k} +mn^{l-ak} - mn^{l-(a-1)k} +mn^{l-(a-2)k} + \dots + (-1)^{a} mn^{l-k} +(-1)^{a+ 1} =0 $ If $m \geq 2$ then $n^{l-(a-1)k} +m(n^{l-ak} - n^{l-(a-1)k} +n^{l-(a-2)k} + \dots + (-1)^{a} n^{l-k}) +(-1)^{a+ 1} < n^{l-(a-1)k} +2(n^{l-ak} - n^{l-(a-1)k} + \dots + (-1)^{a} n^{l-k}) +(-1)^{a+ 1} = -n^{l-(a-1)k} + 2n^{l-ak} + \dots $

Here's my solution (maybe different): Let $\frac{n^{k+l}-1}{n^k+mn^l+1}=z$. We make two cases as follows (both are quite similar)- First consider $k \geq l$. Then, working modulo $n^l$, we see that $z \equiv -1 \pmod{n^l} \Rightarrow n^l \leq z+1$. However, we have $$n^kz<(n^k+mn^l+1)z=n^{k+l}-1<n^{k+l} \Rightarrow n^l-1 \geq z<n^l \Rightarrow z=n^l-1$$This gives $(n^k+mn^l+1)(n^l-1)=n^{k+l}-1$. On opening the brackets, we get $m=\frac{n^{k-l}-1}{n^l-1}$. As $m \in \mathbb{N}$, so $n^l-1 \mid n^{k-l}-1 \Rightarrow l \mid k-l \Rightarrow l \mid k$. $\text{ }$ Now we take up $k<l$. This time working modulo $n^k$, we have $z \equiv -1 \pmod{n^k}$. Also, in a similar fashion, we get that $z<n^k$, thus giving $z=n^k-1$. Then we have $(n^k+mn^l+1)(n^k-1)=n^{k+l}-1$. On opening the brackets, we see that $n^{2k}+(m-1)n^{k+l}=mn^l$. This equation has no solutions when $2k>l$. So we must have $2k \leq l$. Suppose that $l>2k$. Dividing the equation by $n^{2k}$, we get that $$m=\frac{n^{l-k}-1}{n^{l-k}-n^{l-2k}} \Rightarrow n^{l-k}-n^{l-2k} \mid n^{l-k}-1 \Rightarrow n \mid n^{l-k}-n^{l-2k} \mid n^{l-2k}-1 \Rightarrow n \mid 1 \rightarrow \text{CONTRADICTION}$$So we must have $l=2k$, which directly gives $m=1$, thus proving the result. $\blacksquare$

First we have $n^k + mn^\ell + 1\mid n^{k+\ell} + n^k + mn^\ell$. If $\ell \le k$, then $n^k + mn^\ell + 1\mid n^\ell(n^k+n^{k-\ell}+m)$, but since $n^\ell$ is relatively prime to the LHS, we can reduce this to $n^k+mn^\ell+1\mid n^k+n^{k-\ell}+m$. This is equivalent to $n^k+mn^\ell+1\mid n^{k-\ell} + m - mn^\ell-1$. Note that $n^{k-\ell} + m < n^k+mn^\ell+1$ and $mn^\ell+1 \le n^k+mn^\ell+1$, and each is positive, their difference is certainly less than $n^k+mn^\ell+1$. Therefore, $n^{k-\ell} + m -mn^\ell-1=0$, so $m = \tfrac{n^{k-\ell}-1}{n^\ell-1}$. Since this is an integer, $\ell\mid k-\ell$, so $\ell \mid k$. If $\ell > k$, then $n^k + mn^\ell+1\mid n^k(n^\ell+mn^{\ell-k}+1)$. Since $n^k$ is relatively prime to the LHS, $mn^\ell+n^k+ 1\mid n^\ell+mn^{\ell-k}+1$. If $m\ge 2$, then clearly the LHS is larger than the RHS, and the RHS is positive, contradiction. So $m=1$. Then $n^\ell + n^k + 1\mid n^\ell + n^{\ell-k} +1$. So $n^\ell+n^k+1\mid n^{\ell-k} - n^k$. But now the LHS is greater than the RHS (again!), so the RHS is 0. Therefore, $\ell=2k$.

Here us nice bounding solution for $k \ge l$, which I found together with @blastoor. Note that \begin{align*} n^l(n^k + mn^l+1) - (n^{k+l}-1)=\\ =n^{k+l}+mn^{2l} + n^l - n^{k+l} + 1= \\ = n^l + mn^{2l}+1\\ \end{align*}We conclude that: $$n^k + mn^l+1 \mid n^l + mn^{2l}+1$$As a result: \begin{align*} n^l + mn^{2l}+1 \ge n^k + mn^l+1 \\ mn^l(n^l -1) \ge n^l(n^{k-l}-1) \\ m(n^l-1) \ge n^{k-l} -1 \\ m \ge \frac{n^{k-l} -1}{n^l -1} \end{align*}On the other hand note that: $$ n^k + mn^l+1 \mid n^k + mn^l+1+n^{k+l} -1 = n^l(n^{k-l} +m+n^k)$$$$ n^k + mn^l+1 \mid n^{k-l} +m+n^k$$Therefore: \begin{align*} n^{k-l} +m+n^k \ge n^k + mn^l+1\\ n^{k-l}-1 \ge m(n^l-1) \\ \frac{n^{k-l}-1}{n^l-1} \ge m \end{align*}We conclude that: $$ \frac{n^{k-l}-1}{n^l-1} \ge m \ge \frac{n^{k-l}-1}{n^l-1} \implies m = \frac{n^{k-l}-1}{n^l-1}$$This implies that $l \mid k-l \implies l \mid k$ as desired. For case when $l > k$ w can proceed similarly as @v_Enhance did, so I will not elaborate this case.

Let $A=n^k+mn^l+1$. We split the problem into two cases: Case 1: $l > k$ From the problem's condition, we have that $n^{k+l} \equiv 1 \pmod{A}$ $(*)$ Then, observe that from $(*)$, $$ 0 \equiv n^{2k}+mn^{k+l}+n^k \equiv n^{2k}+n^k+m \pmod{A} (**) $$Hence, $$ 0 < n^k+n^l+m \leq n^k+mn^l+1=A \leq n^{2k}+n^k+m$$$\implies 2k \geq l$ $(***)$ On the other hand, from $(*)$, we have that $$ 0 \equiv n^{k+l}+mn^{2l}+n^l \equiv mn^{2l}+n^l+1 \equiv mn^{l-k}+n^l+1 \pmod{A}$$Thus, from $(***)$, $$n^k+mn^l \leq mn^{l-k}+n^l \implies 1+mn^{l-k} \leq mn^{n-2k}+n^{l-k} \leq m+ n^{l-k}$$$\implies n^{l-k}(m-1) \leq m-1 \implies m=1$, since $n >1$. Then, from $(**)$, we have that $$n^k+n^l+1|n^{2k}+n^k+1$$$\implies n^k+n^l+1|n^{2k}-n^l=n^l(n^{2k-l}-1) \implies n^k+n^l+1|n^{2k-l}-1$. Since $|n^k+n^l+1| > |n^{2k-l}-1|$, $n^{2k-l}-1=0 \implies 2k=l$, as desired. $\square$ Case 2: $l \leq k$ From $(*)$, we have that $$ 0 \equiv n^{2k}+n^k+m \equiv n^{k-l}+n^k+m \pmod{A} $$$\implies A|n^{k-l}+n^k+m-A=n^{k-l}-mn^l+m-1$, but since $|A| > |n^{k-l}-mn^l+m-1|$, we have that $n^{k-l}-mn^l+m-1=0 \implies m=\frac{n^{k-l}-1}{n^l-1}$, and this implies that $l|k-l \implies l|k$. Hence, we are done. $\blacksquare$

IMO 2016 SL N4 wrote: Let $n, m, k$ and $l$ be positive integers with $n \neq 1$ such that $n^k + mn^l + 1$ divides $n^{k+l} - 1$. Prove that $m = 1$ and $l = 2k$; or $l|k$ and $m = \frac{n^{k-l}-1}{n^l-1}$. Suppose firstly that $k=\ell$. Then, $(n^k+m \cdot n^k+1) \mid (n^{2k}-1)$, hence $$(n^k+m \cdot n^k+1) \mid (n^{2k}+m \cdot n^k+n^k)=n^k(n^k+m+1)$$Since $\gcd (n^k+m\cdot n^k+1,n^k)=1$, we obtain $(n^k+m \cdot n^k+1) \mid (n^k+m+1)$, which is impossibe since $n >1$ implies $n^k+m \cdot n^k+1 >n^k+m+1$. Now, we split in two cases. Case 1: $k>\ell$. Then we may write $k=\ell+s$ with $s$ a positive integer. Note that $(n^k+m \cdot n^\ell+1) \mid (n^{k+\ell}-1)$ implies $$(n^k+m \cdot n^\ell+1) \mid (n^{k+\ell}+n^k+m \cdot n^{\ell})=n^\ell(n^{\ell+s}+n^s+m),$$therefore we obtain $(n^{\ell+s}+m \cdot n^\ell+1) \mid (n^{\ell+s}+n^s+m)$. Let $n^{\ell+s}+m \cdot n^\ell+1$ and $ n^{\ell+s}+n^s+m=B$. Then $A \mid B$, and $$2A-B=n^{\ell+s}-n^s+2m \cdot n^\ell-m+2>0,$$so $2A >B$. Therefore we conclude that actually $A=B$, which implies that $m=\frac{n^s-1}{n^{\ell}-1}$. We only need to show that $\ell \mid k$, or equivalently that $\ell \mid s$, and we would have the full needed description. But this is almost trivial. Suppose that $\ell \nmid s$, and let $s=X\ell +Y$ with $Y$ being non-zero and $Y<\ell$. Then $$1 \equiv n^s=n^{X\ell+Y}=(n^\ell)^X \cdot n^Y \equiv n^Y \pmod {n^\ell-1}$$therefore $(n^\ell-1) \mid (n^Y-1)$, and since $n^Y-1 \neq 0$, we obtain $Y \geq \ell$, a contradiction. Thus, $ \ell \mid s \Rightarrow \ell \mid k$, and we conclude. Case 2: $\ell>k$. Similarly, let $\ell=k+t$ with $k$ a positive integer. As before, $$(n^k+m \cdot n^\ell+1) \mid (n^{k+\ell}+n^k+m \cdot n^\ell=n^k(n^{k+t}+m \cdot n^t+1),$$hence $(n^k+m \cdot n^\ell+1) \mid (n^{k+t}+m \cdot n^t+1)$. Therefore, $$n^k+m \cdot n^{k+t}+1 \leq n^{k+t}+m \cdot n^t+1 \Rightarrow n^{k+t}-n^k \geq m(n^{k+t}-n^t) \geq n^{k+t}-n^t \Rightarrow n^t \geq n^k,$$thus $t \geq k$. In addition, $$(n^k+m \cdot n^{k+t}+1) \mid (n^{2k+t}-1) \mid (m \cdot n^{2k+t}-m),$$and since $$(n^k+m \cdot n^{k+t}+1) \mid (n^{2k}+m \cdot n^{2k+t}+n^k),$$we obtain $$(n^k+m \cdot n^{k+t}+1) \mid (n^{2k}+m \cdot n^{2k+t}+n^k)-(m \cdot n^{2k+t}-m)=n^2k+n^k+m$$Hence, $$n^k+m \cdot n^{k+t}+1 \leq n^{2k}+n^k+m \Rightarrow n^{2k}-1 \geq m(n^{k+t}-1) \geq n^{k+t}-1 \geq n^{2k}-1,$$since $t \geq k$, as established before. Thus, equality must hold everywhere; that is, $m=1$ and $t=k$, so $m=1$ and $\ell=2k$, hence we obtain the desired description, so we are done in this case, too.

Such a nice looking problem with $4$ variables turns out to be just another easy standard problem . Solved with Pujnk We have $$n^k + mn^l + 1 | n^{k+l} - 1$$ $$\implies n^k + mn^l + 1 | n^{k+l} + n^k + mn^l$$Call this $(1)$ Now, we consider cases. Case 1: $k \ge l$ From $(1)$, we have $n^k + mn^l + 1 | n^l(n^k + n^{k-l} + m)$ $\implies n^k + mn^l + 1 | n^k + n^{k-l} + m$ $\implies m \le \frac{n^{k-l}-1}{n^l - 1}$ However, on the other hand, we have $n^k + mn^l + 1 | n^{k+l} + mn^{2l} + n^l \implies n^k + mn^l + 1 | mn^{2l} + n^l + 1$ $\implies m \ge \frac{n^{k-l}-1}{n^l - 1}$ Which together force $m = \frac{n^{k-l}-1}{n^l - 1}$. Since $m$ is an integer, we must have $n^l - 1 | n^{k-l} - 1 \implies l | k-l \implies l |k$. $\square$ Case 2: $k < l$ From $(1)$, we have $n^k + mn^l + 1 | n^k(n^l + n^{l-k} + 1 \implies n^k + mn^l + 1 | n^l + n^{l-k} + 1$ $\implies m \le \frac{n^{2k-l}(n^{l-k}-1)}{n^k - 1} = \frac{n^k - n^{2k-l}}{n^k - 1} \le 1$. Since $m$ is a positive integer, we must have $m = 1 \implies n^{2k-l} = 1 \implies l = 2k$, which finishes the problem. $\blacksquare$

We will find all of the solutions by casework. Case 1: $k\geq\ell$ Then, $n^k+mn^\ell+1\mid n^{k+\ell}+n^k+mn^\ell$, which means that $n^k+mn^\ell+1\mid n^k+n^{k-\ell}+m$. Therefore, we have $n^k+mn^\ell+1\mid n^{k-\ell}-mn^\ell+m-1$. Since $n^k+mn^\ell>n^{k-\ell}+mn^\ell$ and $n^k+mn^\ell+1>mn^\ell+1$, this means that the only possibility is if $n^{k-\ell}-mn^\ell+m-1=0$. This is equivalent to $$n^{k-\ell}-1=m(n^\ell-1).$$Therefore, we must have $\ell\mid k$. Let $k=(a+1)\ell$. Then, $m=\frac{n^{a\ell}-1}{n^\ell-1}$, so we have the solution $(n,m,k,\ell)=\left(n,\frac{n^{a\ell}-1}{n^\ell-1},(a+1)\ell,\ell\right)$. Case 2: $\ell>2k$ Then, $n^k+mn^\ell+1\mid n^{k+\ell}+n^k+mn^\ell$, which means that $n^k+mn^\ell+1\mid n^\ell+1+mn^{\ell-k}\mid mn^\ell+m+m^2n^{\ell-k}$. Therefore, we have $$mn^\ell+n^k+1\mid m^2n^{\ell-k}-n^k+m-1.$$We have $n^k+1<mn^\ell+n^k+1$, so the right hand side cannot be negative. Suppose that it is positive. Then, $m^2n^{\ell-k}+m>mn^\ell+n^k$, so $mn^{\ell-k}+1>n^\ell$, which means that $mn^{\ell-k}\geq n^\ell$. Therefore, $m\geq n^k$. However, this is impossible since in the original equation, we must have $n^k+mn^\ell+1>n^{k+\ell}$, and $n\neq1$. Therefore, we must have $m^2n^{\ell-k}-n^k+m-1=0$. However, we have $m^2n^{\ell-k}>n^k$, so the expression cannot equal $0$. Case 3: $k<\ell\leq2k$ Then, $n^k+mn^\ell+1\mid n^{k+\ell}+n^k+mn^\ell$, which means that $n^k+mn^\ell+1\mid n^\ell+1+mn^{\ell-k}$, so $n^k+mn^\ell+1\mid n^\ell+mn^{\ell-k}-n^k- mn^\ell$. Therefore, we have $$n^k+mn^\ell+1\mid n^k-n^{2k-\ell}+m-mn^k.$$Since $n^k+mn^\ell+1>n^k+m$ and $n^k+mn^\ell>mn^k+n^{2k-\ell}$, this means that the right hand side must be zero. Therefore, $m(1-n^k)=n^{2k-\ell}-n^k$, so $m=\frac{n^{2k-\ell}-n^k}{1-n^k}$. Since $m$ is an integer, we must have $1-n^k\mid 1-n^{\ell-k}$, so $k\mid\ell-k$. This means that $k\mid\ell$. This means that $\ell=2k$, which gives the solution $(n,m,k,\ell)=(n,1,k,2k)$. Therefore, the only solutions are $\boxed{(n,m,k,\ell)=\left(n,\frac{n^{a\ell}-1}{n^\ell-1},(a+1)\ell,\ell\right)}$ and $\boxed{(n,m,k,\ell)=(n,1,k,2k)}$.

Euclid is cool. \textbf{Solution: } We do case work. Consider first if $l\leq k$. Then, \[n^k+mn^l+1 \mid (n^l-1)(n^k+mn^l + 1) - (n^{k+l}-1) = -n^k+n^l +m n^l (n^l-1) \]Note that $n$ is relatively prime to the LHS, so we can divide through and $n^k+mn^l +1 \mid -n^{k-l} + 1 + m(n^l-1)$. Note that this is clearly less than $n^k+mn^l+1$ and also nonnegative, so we must have $m = \frac{n^{k-l}-1}{n^l-1}$ with $l\mid k$ in order for $m$ to be an integer. Next, consider $l>k$. Then, \[n^k+mn^l+1\mid n^{k+l} + n^k + mn^l = n^k (n^{l}+1 + mn^{l-k})\]If $m=1$, this gives $n^k+n^l+1 \mid n^l+1+n^{l-k}$, and thus $n^l+n^k+1 \mid n^{l-k}-n^k$. The LHS is larger in magnitude than the RHS, so $n^{l-k}=n^k$ and thus $l=2k$ whenever $m=1$. If $m\geq 2$, then $n^k+mn^l + 1 > n^l + mn^{l-k} + 1$ since $m(n^l-n^{l-k}) = n^{l-k} (m(n^k-1))\geq n^{l-k} n^k > n^l - n^k$. Since the RHS is still positive, this is a contradiction so this case does not contribute anything.

Weird expressions in divisibility screams "inequalities", spamming them solves the problem $$n^k + mn^l + 1| n^{k+l}-1$$Rewrite the divisibility as $$ n^k+mn^l+1 | n^{k+l}+n^k+mn^l$$Notice that $\gcd (n^k,mn^l+1,n)=1$, which means we can further simplify RHS, but in order to do that, we will divide into cases. Case 1. $l >k$ Then $$n^k+mn^l+1 | n^l+1+mn^{l-k}$$ Claim 1. $m \le 1$ By divisibility, $$n^k+mn^l+1 \le n^l +1+ mn^{l-k} \Rightarrow m \le \frac{n^l-n^k}{n^l-n^{l-k}} $$. FTSC, assume $ m \ge 2$. Then from our first inequality \begin{align*} &2n^l-2n^{l-k} \le n^l-n^k \leftrightarrow n^l+n^k \le 2n^{l-k} \Rightarrow\\ & \frac{n^k ( n^{l-k} +1)}{n^{l-k}} = n^k+ \frac{1}{n^{l-k}} \le 2. \end{align*}Which is absurd since $\min \{n^k \}=2$. $\square$ Claim 2. $m=1$ works and $l=2k$ iff $m=1$ Let $m=1$. We have \begin{align*} & n^k+n^l \le n^l+n^{l-k} \Leftrightarrow n^{l-k}-n^k \ge 0 \Rightarrow l \ge 2k \qquad (\spadesuit)\\ & \qquad \\ &n^k+n^l+1 | n^l+1 + n^k + n^{l-k} -n^k \Rightarrow \\ & n^k+n^l+1 | n^{l-k} - n^k \stackrel{ (\spadesuit)}{=} n^k( n^{l-2k}-1) \end{align*}Since $\gcd(n^k+n^l+1,n)=1$, we have $n^k+n^l+1 | n^{l-2k}-1$. Clearly, RHS is positive and is smaller than LHS, then $n^{l-2k}-1=0 \rightarrow l=2k$. $\square$ Case 2. $ l \le k$ simplify the divisibility equation as $$ n^k+mn^l+1 | n^k+n^{k-l}+m$$By divisibility inequality \begin{align*} & mn^l+1 \le n^{k-l}+m \Rightarrow m \le \frac{n^{k-l}-1}{n^l-1} \ \text{and}\ k\ge l \qquad (\clubsuit) \end{align*}Notice that $(\clubsuit)$ is the expression we wanted, and this expression is an integer iff $l|k$ ( we will prove later), so, we are in right path \begin{align*}&n^k+mn^l+1 | n^k+n^{k-l}+m - (n^k+mn^l+1) = n^{k-l}+m-mn^l-1 \Rightarrow \\ &n^k+ mn^l+1 | n^{k-l}+m-mn^l-1 \ge 0 \end{align*}Notice that RHS is smaller than LHS, forcing RHS be equal to $0$, hence $$n^{k-l}+m-mn^l-1=0 \Rightarrow m= \frac{n^{k-l}-1}{n^l-1} $$ Claim 3. $\frac{n^{k-l}-1}{n^l-1}$ is an integer if $l|k$ This is actually very well known problem , which is $\gcd( n^a-1,n^b-1) = n^{\gcd(a,b)}-1$, which is direct consequence of repeatedly applied euclidean algorithm, hence we have $\gcd(k-l,l)= \gcd (k,l)=l$, which implies $l|k$ since $k \ge l$. $\square$ Therefore, we have proven our desired claims, hence we are done!

Note that \[n^k+mn^l+1\mid n^{k+l}-1 + n^k+mn^l+1=n^{k+l}+n^k+mn^l\]If $k\le l$ then we have $n^k+mn^l+1\mid n^l+mn^{l-k}+1$ so $n^k+mn^l+1\le n^l+mn^{l-k}+1$. Thus, we have \[(m-1)n^l\le n^l-n^k < n^l\]so $m=1$. We have $n^k+n^l+1\mid n^l+n^{l-k}+1 - (n^k+n^l+1)=n^{l-k}-n^k$. Clearly, $n^k+n^l+1> |n^{l-k}-n^k|$, so $n^k=n^{l-k}$, implying $l=2k$, as desired. If $k>l$ then $n^k+mn^l+1\mid n^k+n^{k-l}+m$. This implies that $mn^l+1\le n^{k-l}+m$, so \[m\le \frac{n^{k-l}-1}{n^l-1}\]However, $n^{k+l}-1<n^l(n^k+mn^l+1)$ so \[n^{k+l}-1\le (n^l-1)(n^k+mn^l+1)=n^{k+l}+mn^{2l}+n^l-n^k-mn^l-1\]but this implies that \[m\ge \frac{n^{k-l}-1}{n^l-1}\]which implies that they're equal. Naturally, this implies that $l\mid k$ and we are done.

Note $n^k + mn^l +1$ divides $ n^l\bigl(n^k + mn^l +1\bigr) - \bigl(n^{k+l} -1\bigr) = mn^{2l}+n^l+1$ and $n^k\bigl(n^k + mn^l +1\bigr) - \bigl(n^{k+l} -1\bigr) =n^{2k}+n^k+m $. Hence we get $$n^k + mn^l +1 \leqslant mn^{2l}+n^l+1 \dots [\clubsuit] \text{ and } n^k + mn^l + 1 \leqslant n^{2k}+n^k+m \dots [\spadesuit]$$Now if $(n^k + mn^l +1 )\cdot t = n^{k+l} -1$, then going modulo $n^{\min\{k,l\}} : t \equiv -1 $ mod $n^{\min\{k,l\}}$. Hence $ t\geqslant n^{\min\{k,l\}} -1$. Hence $$(n^k + mn^l +1 )\cdot (n^{\min\{k,l\}}-1) \leqslant n^{k+l} -1 \dots [\diamondsuit]$$We now split into two cases : Case 1 : $k \geqslant l$. Expanding $[\diamondsuit] : mn^{2l} + n^l +1\leqslant n^k + mn^{l} +1$. Combining with $[\clubsuit]$, we must have that equality holds. This implies $m= \frac{n^{k-l}-1}{n^l-1}$. This will be integer only when $l \mid k-l \implies l \mid k$. This indeed fits and gives us our first solution. Case 2 : $k < l$. First from $[\spadesuit] : m \leqslant \frac{n^{2k}-1}{n^l-1} \implies 2k \geqslant l$. Now expanding $[\diamondsuit] : n^{2k} + mn^{l+k} \leqslant mn^l + n^{k+l}$. Now dividing throughout by $n^k$ and some rearranging gives $m \leqslant \frac{n^k - n^{2k-l}}{n^k -1} \overset{2k \geqslant l}{\leqslant} \frac{n^k - 1}{n^k-1} =1$. As $m\geqslant 1$, we must have all inequalities in last chain to be equalities, implying $m=1$ and $2k-l=0$. This indeed fits and gives us our second solution! Hence, we are done!

First we may rewrite the divisibility as \[ n^k + mn^l + 1 \mid n^{k+l} + n^k + mn^l\] If $k\ge l$, then since $n$ is relatively prime to the LHS, we may divide out $n^l$ in the RHS and get \[n^k + mn^l + 1 \mid n^k + n^{k-l} + m\implies n^k + mn^l + 1\mid n^{k-l} + m - mn^l - 1\]Clearly $|n^{k-l} + m - mn^l - 1| \le n^k + mn^l + 1$, so $n^{k-l} + m - mn^l - 1 = 0$, which means $m = \frac{n^{k-1} - 1}{n^l - 1}$, but since $m$ is an integer, we must have $l\mid k-l\implies l\mid k$ . If $k< l $, then we may divide out $n^k$ in the RHS and get \[n^k + mn^l + 1 \mid n^l + mn^{l-k} + 1,\]and thus $n^l + mn^{l-k} \ge n^k + mn^l$. If $m>1$, we have the following: \begin{align*} n^l + mn^{l-k} - n^k - mn^l \\ = m n^{l-k} - (m-1)n^l - n^k \\ = (m - n^k (m-1)) n^{l-k} \\ < (m - 2(m-1)) n^{l-k} \\ < 0,\\ \end{align*}so $n^k + mn^l > n^l + mn^{l-k}$. If $m=1$, then $n^k + n^l + 1\mid n^l + n^{l-k} + 1$, so $n^k + n^l + 1 \mid n^{l-k} - n^{k} $. Now, we have that $n^l$ is greater than both $n^{l-k}$ and $n^k$, we have $|n^k + n^l + 1| > |n^{l-k} - n^k|$, so $n^{l-k} - n^k = 0$, which implies $l = 2k$, as desired.

Consider two cases. Case 1: $l \leq k$. Note that $$n^k+mn^l+1 \mid n^{k+l}-1-(n^k+mn^l+1)(n^l-1) = n^l(n^{k-l}+m-mn^l-1)$$ Since $\gcd(n^k+mn^l+1,n^l) = 1$ we have $$n^k+mn^l+1 \mid n^{k-l}+m-(mn^l+1)$$ However, since $n^{k-l}+m < n^k+mn^l+1$ and $mn^l+1 < n^k+mn^l+1$ it follows that $n^k+mn^l+1 > |n^{k-l}+m-(mn^l+1)|$ which means that $$n^{k-l}+m -(mn^l+1) = 0$$from which we extract $$m = \frac{n^{k-l}-1}{n^l-1}$$which is only an integer if $l \mid k-l$ or $l \mid k$ as desired. Case 2: $l > k$. In this case, note that $$n^k+mn^l+1 \mid n^{k+l}-1+(n^k+mn^l+1) = n^k(n^l+mn^{l-k}+1)$$which implies $$n^k+mn^l+1 \mid n^l+mn^{l-k}+1$$since $\gcd(n^k+mn^l+1, n^k) = 1$. However, if $m > 1$ then $n^k+mn^l+1 > n^l+mn^{l-k}+1$ because $n^l(m-1) > mn^{l-k}$ since $n^k \geq 2$. But $n^l+mn^{l-k}+1$ can't be $0$, so $m = 1$. If $m = 1$, then $$n^k+n^l+1 \mid n^{l-k}+n^l - (n^k+n^l+1) = n^{l-k}-n^k$$but clearly $|n^{l-k}-n^k| < n^k+n^l+1$ so $n^{l-k} = n^k$ which directly gives us $l = 2k$ as desired. Actually, it seems that simply subtracting the LHS from the RHS can be the first step for both cases but whatever.

Pretty easy problem for an N4, but maybe that's because I've been recently consistently solving N4-5s idts tho We claim the answer is $m = 1$ with $l = 2k$, or $l\mid k$ with $m = \frac{n^{k-l}-1}{n^l-1}$ (the version I received wanted me to find the sol set). Case 1. $l\le k$. It's equivalent to $n^k+mn^l+1\mid n^{k+l}+n^k+mn^l\implies n^k+mn^l+1\mid n^k+n^{k-l}+m-n^k-mn^l-1=n^{k-l}-mn^l+m-1$, but by caseworking on what the sign of the RHS is, we find that the absolute value of RHS<LHS, meaning RHS=0; in particular, solving, we get $m=\frac{n^{k-l}-1}{n^l-1}\rightarrow l\mid k-l\rightarrow l\mid k$. Case 2. $l>k$. Similarly we get $LHS\mid n^l+mn^{l-k}+1$; supposing $m\ge2\implies LHS=n^k(1+mn^{l-k})+1>n^k(n^{l-k}+mn^{l-2k})+1=RHS>0$, contradiction (easy to verify since $1+(m-1)n^{l-k}$ increases faster than $mn^{l-2k}$), hence m=1. We conclude that $n^k+n^l+1\mid n^l+n^{l-k}+1\implies LHS\mid n^{l-k}-n^k$, again since RHS<LHS in this expression we must have $n^{l-k}-n^k=0\implies l=2k$.

Note that this is equivalent to \[ n^k + mn^\ell + 1 \mid n^{k+\ell} + n^k + mn^\ell. \]and that the LHS is coprime with $n$. Claim: If $\ell \le k$, then $m = \frac{n^{k-\ell} - 1}{n^{\ell} - 1}$ for $\ell \mid k$ is one solution set. Proof. This simplifies as \[ n^k + mn^\ell + 1 \mid n^{k} + n^{k-\ell} + m. \]which becomes \[ n^k + mn^\ell + 1 \mid n^{k-\ell} + m - mn^\ell - 1. \]Since the RHS has smaller absoluted value, it must follow that it equals zero, giving the result. $\blacksquare$ Claim: If $\ell > k$, then $m = 1$ and $\ell = 2k$ is another solution set. Proof. The expression then otherwise simplifies as \[ n^k + mn^\ell + 1 \mid n^{\ell} + mn^{\ell - k} + 1 \]If $m \ge 2$, the LHS becomes too large, giving a contradiction. As such, $m = 1$ and $n^\ell + n^k + 1 \mid n^{\ell} + n^{\ell - k} + 1$, which only holds if $\ell = 2k$. $\blacksquare$

Pretty nice problem, though easy for an N4. The problem condition is equivalent to $n^k + mn^l + 1 \mid n^{k+l} + mn^l + n^k$. Now consider the two cases: Case 1: $l > k$. Then $n^k + mn^l + 1 \mid n^l + mn^{l-k} + 1$. If $m > 1$, then $n^k(n^{l-k} - 1) < n^k \cdot n^{l-k} < m(n^k - 1)\cdot n^{l-k}$, so $n^k + mn^l + 1 > n^l + mn^{l-k} + 2$, a contradiction. Thus $m = 1$. Hence we have $n^k + n^l + 1 \mid n^l + n^{l-k} + 1$, so $n^k + n^l + 1 \mid n^{l-k} - n^k$ and since $n^{l-k} - n^k < n^l + n^k + 1$, therefore $n^{l-k} = n^k$ implies $l = 2k$. Case 2: $l \le k$. In this case, we have $n^k + mn^l + 1 \mid n^k + m + n^{k-l}$. Assume $n^k + m + n^{k-l} > n^k + mn^l + 1$. Then this yields $n^k + mn^l + 1 < n^k + m + n^{k-l} - n^k - mn^l - 1 = (n^{k-l} - 1) - m(n^l - 1) < n^{k-l} - 1 < n^k$, a contradiction. Therefore $n^k + m + n^{k-l} = n^k + mn^l + 1$, so $l \mid k$ and $m = \frac{n^{k-l} - 1}{n^l - 1}$.

I find it very superfluous for the problem itself to give the explicit curve; it's much more natural to characterize, say, all quadruples that work. Rewrite the given condition as $$n^k +mn^\ell + 1 \mid n^k + n^{k+\ell}+mn^\ell.$$The divisor will remain fixed throughout the solution, say at $S$. First, suppose that $\ell \leq k$. Then as $S \mid n^{k-\ell} + n^k + m$, we have $S \mid n^{k-\ell} - mn^\ell -1 + m$. In particular, notice that $$n^{k-\ell} - mn^\ell - 1 < n^{k-\ell} + mn^\ell - 1 < S,$$so we must have $m(n^\ell - 1) = n^{k-\ell} - 1$, which yields the first curve. Next, if $\ell > k$, then the condition reduces to $S \mid mn^{\ell - k} + n^\ell + 1$. If $m \geq 2$, then the RHS is smaller than $S$; hence, $m = 1$, and in particular $$n^k + n^\ell + 1 \mid n^{\ell-k}+n^\ell + 1.$$This simplifies to $n^k+n^\ell + 1 \mid n^{\ell-2k} - 1$, which implies that $\ell = 2k$, the second solution curve.

Standard Size Arguments $\textbf{Case I:}$ $k\geq l$ We have \begin{align*} &n^k+mn^l+1\mid n^{k+l}-1-n^k-mn^l-1=n^{k+l}-n^k-mn^l\\ \implies & n^k+mn^l+1\mid n^k+n^{k-l}+m\\ \implies & n^k+mn^l+1\mid n^{k-l}+m-mn^l-1 \end{align*}The last part is simply not possible because $n^{k-l}-1-m(n^l-1)<n^{k-l}-1$ this give us $n^{k-l}-1-m(n^l-1)=0\implies n^{k-l}-1=m(n^l-1)\implies l\mid k$. $\textbf{Case II:}$ $l>k$ \begin{align*} &n^k+mn^l+1\mid n^{k+l}-n^k-mn^l\\ \implies &n^k+mn^l+1\mid n^l+mn^{l-k}+1 \end{align*}Putting $m=1$ we get $n^k+n^l+1\mid n^l+n^{l-k}+1\iff n^k+n^l+1\mid n^{l-k}-n^k $, but this gives us $LHS>RHS$, so $n^{l-k}=n^k\implies \boxed{l=2k}$