The answer is no; suppose, for the sake of contradiction, that such a function $f$ exists. Let $P(m,n)$ be the assertion in the problem. Notice that for any two constants $a$ and $b$, the assertions \[P(a, b+f(a+bf(a))) \text{ and } P(a+bf(a), f(a))\]have equal left hand sides. So, we may equate the right hand sides: \begin{align*} f(f(a))^{a+bf(a)} + 2024! \cdot (a+bf(a)) &= f(\text{blah})^{a} + 2024! \cdot a \\ f(f(a))^{a+bf(a)} + 2024! \cdot bf(a) &= f(\text{blah})^{a}.\end{align*}But if $a \geq 3$ and $b = a^a \cdot 2024!^{a-1} \cdot f(a)^{a-1}$, the equation is false by Fermat's last theorem.

Overcomplicated NT sledgehammer oops. No, such a function does not exist. Let $k=f(1)$ throughout. The first step of this solution is to construct an arithmetic progression of $f$ with the desirable property. Claim. There exists some prime $p$ and a sequence $a_1,a_2,\dots$ such that $p$ divides all terms $a_i$ and $f(a_1), f(a_2), \dots$ form an arithmetic progression. Proof. First, we note that by plugging in $m=1$, $$f(nk+1) = f(n)+2024!.$$Thus, if we define a sequence $b_1,b_2,\dots$ by $b_1=1$ and $b_{i+1} =kb_i+1$, we get that $f(b_1), f(b_2),\dots$ is an arithmetic progression. We now prune this sequence so that every term is divisible by $p$. Pick a prime $p\mid k+1$. Then, the sequence $b_1,b_2,\dots$ cycles modulo $p$. Thus, the terms that are divisible by $p$ are equally spaced, and we can pick those as a subsequence. $\blacksquare$ Claim. There exists $A$ and $B$ such that $\{f(An+B)\}_{n\in\mathbb Z^+}$ is bounded. Proof. Take $a_1,a_2,\dots$ from the previous claim, and let $d=\gcd(f(a_1),f(a_2),\dots)$. By Dirichlet's theorem, there exists a sequence $x_1<x_2<\dots$ such that $f(a_{x_i})$ is $d$ times a prime number. By pigeonhole, there exists $i$ and $j$ such that $d\mid a_{x_i} - a_{x_j}$. Let $u = a_{x_i}$ and $v = a_{x_j}$. We may assume $u,v > 3$. Notice that since $d\mid u-v$, but $\gcd(f(u), f(v))=d$, by the Chinese remainder theorem, there exists $n_0$ such that \begin{align*} n_0 &\equiv u\pmod{f(u)} \\ n_0 &\equiv v\pmod{f(v)}. \end{align*}Therefore, we have that $$f\left(\frac{n_0-u}{f(u)}\right)^u + 2024!\cdot u = f(n_0) = f\left(\frac{n_0-v}{f(v)}\right)^v + 2024!\cdot v.$$Moreover, we may replace $n_0$ above with $n = n_0+tf(u)f(v)$ for any $t$, giving $$f\left(\frac{n_0-u}{f(u)} + tf(v)\right)^u + 2024!\cdot u = f\left(\frac{n_0-v}{f(v)} + tf(u)\right)^v + 2024!\cdot v.$$However, from the condition in the first claim, we have $u$ and $v$ divisible by some prime $p$. Therefore, we have the difference of two perfect $p$-th power is a constant, which has finitely many solutions. Thus, $A=f(v)$ and $B=\tfrac{n_0-u}{f(u)}$ works. $\blacksquare$ To finish, note that $f(m+Af(m)) = f(A)^m + 2024! m$. Now, let $m = Ai+B$, where $i\in\mathbb Z^+$ varies. The left-hand side is bounded by the claim, but the right-hand side is not, a contradiction.

Here is the size argument (without FLT) I presented during test review . Note that for any choice $a, k, l\in \mathbb{Z}_{>0}$ the quadruple $$a=a, \quad b= k+lf(a+f(a)k), \quad c=a+f(a)k, \quad d=f(a)l$$satisfies $a+bf(a)=c+df(c)$. From the condition it then follows that $$f(k+lf(a+f(a)k))^a - f(f(a)l)^{a+f(a)k} = 2024!f(a) k.$$ Claim: For any $a>1$, $\{f(f(a)l)\}_{l>0}$ is bounded. Proof: Fix $k=a$ in the above. Then the LHS is a difference of $a$'th powers as $l$ varies, while the RHS is positive and bounded. This is only possible if the claim is true. $\blacksquare$ However, setting $m=f(a)p$, $n=f(a)$ in the original equation we find that $$f(f(a))^{f(a)p} + 2024!f(a)p = f(f(a)(p+f(f(a)p))$$is bounded. Sending $p\to \infty$ yields a contradiction. Hence no such $f$ exists.

Sketch: for all $p\nmid2024!$: Suppose $\gcd(k,p)=1$. $f(kx)$ is surjective modulo $p$ for $k$ fixed $f(kx+c)$ is surjective modulo $p$ for $k$ and $c$ fixed Now finish by $m=2$ and take a NQR modulo $p$

Beautiful problem!

The answer is no. Let $p \nmid 2024!$ be a prime. Fix $m=k(p-1)$ and let $n$ vary; we find that if $x \equiv k(p-1) \pmod{f(k(p-1))}$, then $f(x)$ is one of two values (dependent on $k$) modulo $p$. Now let $n=f(k(p-1))$ and vary $m \equiv k(p-1) \pmod{f(k(p-1))(p-1)}$. Since $m+nf(m) \equiv k(p-1) \pmod{f(k(p-1))}$, as $m$ varies the RHS should attain at most $2$ values modulo $p$. Since $f(n)^m$ is fixed modulo $p$ this clearly requires $p \mid f(k(p-1))(p-1) \implies p \mid f(k(p-1))$. Varying $k$, we get $p-1 \mid n \implies p \mid f(n)$. Finally, from $m=n=p-1$ we obtain a contradiction, since the assertion implies $0 \equiv 0-2024! \pmod{p}$. $\blacksquare$

Very cute. Here's a short proof that uses simple bounding. The answer is $\boxed{\text{no}}$. Assume the contrary. Let $P(m,n)$ denote the given FE. Let $c=2024!$. Claim: If $f(2) \mid k$ then $f(k)=1$. Proof: Choose a large even positive integer $m$ such that $f(2) \mid m-2$. Let $n=\frac{m-2}{f(2)}+\frac{k}{f(2)}f(m)$, so that $2+nf(2)=m+kf(m)$. Now, comparing $P(2,n)$ and $P(m,k)$, we get $$c(m-2)=f(n)^2-f(k)^m=(f(n)+f(k)^{\frac{m}{2}})(f(n)-f(k)^{\frac{m}{2}}).$$Since the LHS is a positive integer, so is the RHS, and so the second term in brackets is at least $1$. But the first term in brackets is at least $f(k)^{\frac{m}{2}}$, so we get $$f(k) \leq (c(m-2))^{\frac{2}{m}}.$$As $m \rightarrow \infty$, RHS tends to $1$, so we get $f(k)=1$, as required. $\blacksquare$ Now, take $m_1>m_2$ such that $f(m_1)=f(m_2)=1$. Then, comparing $P(m_1,m_2)$ and $P(m_2,m_1)$ we get $$f(m_1+m_2)=1+cm_1>1+cm_2=f(m_2+m_1).$$Contradiction!

First p3/6/9 solve in contest Denote the assertion by $P(m,n).$ Taking $P(a,n)$ gives $f(a+bn)=f(n)^a+2024!\cdot a,$ where $b=f(a).$ Now taking $P(a,a+bm(f(a)^a+2024!\cdot a))$ and $P(a+ab,b^2m)$ gives $f(a+bm(f(a)^a+2024!\cdot a))^a-(f(b^2m)^{1+b})^a=2024!\cdot ab.$ In particular, the left hand side is a difference of two $a$th powers and is thus $\ge 2^a-1.$ Thus we get $f(a)=b\ge\frac{2^a-1}{2024!\cdot a}.$ Now if $f(1)=c,$ taking $P(1,1)$ gives $f(1+c)=c+2024!.$ Taking $P(1,1+c)$ gives $f(1+c+c^2)=c+2\cdot 2024!,$ and $P(1,1+c+c^2)$ gives $f(1+c+c^2+c^3)=c+3\cdot 2024!.$ Similarly, by induction we get $f(1+c+\dots+c^k)=c+k\cdot 2024!.$ But setting $1+c+\dots+c^k=S,$ we get $S\ge k+1$ and thus $f(S)\ge\frac{2^{k+1}-1}{2024!\cdot(k+1)}$ for large enough $k.$ But as $k$ grows, this lower bound is $O\left(\frac{2^k}k\right),$ and the closed form we get is $O(k),$ contradiction. Thus no function exists.

Here is a solution that uses only 1% of the FE and 99% bounding Let $a\ge 2025!$ be fixed.Also let $k\in \mathbb{Z_+}$ vary. Denote $P(m,n)$ the assertion. We have by $P(a,k+f(a+f(a)k))$: $$f(a+kf(a)+f(a)f(a+f(a)k))=f(k+f(a+f(a)k))^a+2024!\cdot a=\text{something}^a+2024!\cdot a$$But by $P(a+f(a)k,f(a))$ we have that $$f(a+f(a)k+f(a)f(a+f(a)k))=f(f(a))^{a+f(a)k} +2024!\cdot (a+f(a)k)$$Subtracting the equations we get that ($x=f(f(a))$ is constant) $x^{a+f(a)k}+2024!\cdot f(a)k=\text{something}^a$ for every $k$ we divide into cases: Case 1: $x\ge 2$ Let $k=al$ where $l\in \mathbb{Z_+}$.Then $x^{a+alf(a)}+2024!\cdot af(a)l=\text{something}^a$ so $x^{a+alf(a)}+2024!\cdot af(a)l\ge (x^{lf(a)+1}+1)^a$ meaning $2024!\cdot af(a)l\ge x^{(a-1)(f(a)l+1)}\cdot a\ge 2^{(a-1)f(a)l}\cdot a$ and denoting $f(a)l=z$ we get that $2024!\cdot z\ge 2^{(a-1)z}\ge 1+(a-1)z\ge 1+2024!z$ false Case 2: $x=1$ So we have that $2024!\cdot f(a)k+1=\text{something}^a$ but then just choose$ k=(2024!f(a))^{a-1}$ so you have $2$ perfect powers of $a$ with difference $1$ contradiction so no function exists.

Actually it’s easy by mod few primes

There exists no such function $f$. Denote the given equation by $P(m,n)$. By equating the LHS of $P(m_1=a+af(a),n_1=tf(a))$ and $P(m_2=a,n_2=a+tf(a+af(a)))$ we get $$f(n_1)^{m_1}+2024!\cdot m_1=f(a+af(a)+tf(a)f(a+af(a)))=f(n_2)^{m_2}+2024!\cdot m_2.$$(sorry for unreadable latex) Rearranging we get $f(n_1)^{m_1}-f(n_2)^{m_2}=2024!\cdot(m_2-m_1).$ The key observation is that $m_1$ and $m_2$ are both multiples of $a$, so if we choose $a\geq2$, then we force a diophantine equation(!!) of the form $X^a-Y^a=Z$ for some nonzero $Z$ which has finitely many solutions $(X,Y)$. In particular, this means that as $t$ varies over $\mathbb N$, $f(tf(a))$ takes on only finitely many values. But by substituting $P(bf(a),f(a))$ we get $f(bf(a)+f(bf(a))f(a))=f(f(a))^{bf(a)}+2024!\cdot bf(a)$, but the LHS is bounded by the above and the RHS is unbounded by setting $b$ large, a contradiction.

The answer is no. Let $p = 2027$ (yay '27!) and fix an $m$ such that $p-1 \mid m$. Then over all $n \in \mathbb N$, $f(n)^m$ only takes two values modulo $p$, so it follows that $f(m+nf(m))$ only takes two values modulo $p$. In other words, among all positive integers congruent to $m$ mod $f(m)$, $f$ only outputs two distinct values modulo $p$. Now pick a $k\equiv m \pmod {f(m)}$ and substitute $f(m)$ for $n$ in the assertion to obtain \[f(f(m)f(k) + k) = f(f(m))^k + 2024! \cdot k.\]Suppose that $p \nmid f(m)$. Then there exists a residue $r$ and positive integers $k_0, k_1, k_2$ such that $k_i \equiv m \pmod {f(m)}$, $k_i \equiv r \pmod {p-1}$, and $k_i \equiv i \pmod p$ for each $0 \leq i \leq 2$. It follows that $f(f(m)f(k_i) + k_i)$ are distinct modulo $p$, contradiction. It follows that $p \mid f(m)$. As our choice of $m$ was arbitrary, it follows that $p \mid f(m)$ for all $p-1 \mid m$. But now take $m$ and $n$ to both be multiples of $p-1$ in the original equation, such that $p \mid f(n)^m$ and $p \mid f(m+nf(m))$. It follows that $p \mid 2024! \cdot m$, which yields a contradiction if we pick $p \nmid m$, as needed. Remark: The problem is much harder than the solution seems; when mocking the contest, I figured out the first claim rather quickly, but it turns out that interpreting it the right way (``$f$ outputs only two distinct values") to motivate the second substitution is really counterintuitive.