Let $q=m+1\equiv3\pmod{4}$. Then $m\equiv2\pmod{4}$, so $n,2^n-1\ge 2 \implies p^a\equiv \frac{-1}{m-1}\equiv -1\pmod{4}$, whence $p\equiv3\pmod{4}$ and $a$ is odd. Let $0\le v_2(n+1)=r\le n-1$. Then \[m^{2^r}+1 \bigg{|} \frac{m^{2^n}-1}{m-1} - (m^{n+1}+1) = m p^a.\]But $\gcd(m,m^{2^r}+1)=1$, so $m^{2^r}+1=p^b$ for some $1\le b\le a$. If $r\ge1$, then $m^{2^r}\equiv-1\pmod{p}\implies p\equiv 1\pmod{4}$, a contradiction. On the other hand, if $r=0$ (so $n$ is even), then $q=m+1=p^b\implies p=q$ and $b=1$. If $n\ge3$, then \[\frac{-1}{p-2}\equiv \frac{m^{2^n-1}-1}{m-1} = m^n+p^a\equiv p(p^2)^{\frac{a-1}{2}}\equiv p\pmod{8}\implies 8|(p-1)^2,\]contradicting the fact that $p\equiv3\pmod{4}$. Thus $n=2$, which clearly works by taking $a=1$ and $p=q$, so we're done.

math154 wrote: Let $0\le v_2(n+1)=r\le n-1$. Then \[m^{2^r}+1 \bigg{|} \frac{m^{2^n}-1}{m-1} - (m^{n+1}+1) = m p^a.\]. I don't know why maybe I am tired but I cannot prove that

agzqu wrote: Fine all positive integers $m,n\geq 2$, such that (1) $m+1$ is a prime number of type $4k-1$; (2) there is a (positive) prime number $p$ and nonnegative integer $a$, such that \[\frac{m^{2^n-1}-1}{m-1}=m^n+p^a.\] Claim: $n$ is even. Proof: Let $\nu_2(n+1)=r$,then $m^{2^r}+1|mp^a$ and since $\gcd(m^{2^r}+1,m)=1$ so $m^{2^r}+1|p^a$,a contradiction so $n$ must be even. This forces $p=m+1|p^a$ for $p=q$ Now expanding our original equation gives $q+m^2+m^3............+m^{2^n-2}=m^n+q^a$ If $n \ge 3$ then the left hand side is congruent to $q+4 \pmod 8$ and the right hand side is congruent to $q \pmod 8$,a contradiction.Hence $n=2$ and the equation reduces to $m+1=p^a$ or $a=1$ hence we are done.

Nice problem involving NT along with some polynomial manipulations... Note that: $$\frac{m^{2^n-1}-1}{m-1} - m^n = p ^a$$Let $\nu_p(n+1) = k$, $n+1 = 2^k s$ and $D = \frac{m^{2^n-1}-1}{m-1} - m^n$. Claim: $m^{2^k}+1|D$ Proof: Notice that, it suffice to prove $m^{2^k}+1|mD$ as $\gcd(m, m^{2^k}+1) = 1$. Hence: $$mD = \frac{m^{2^n}-1}{m-1}-m^{n+1} = (m+1)(m^2+1) \cdots (m^{2^{n-1}}+1) - (m^{2^k}+1)(m^{2^k(s-1)} - m^{2^k(s-2)} + \cdots) $$$$= (m^{2^k}+1)\left((m+1)(m^2+1) \cdots (m^{2^{r-1}}+1)(m^{2^{r+1}}+1) \cdots (m^{2^{n-1}}+1)) - m^{2^k(s-1)} - m^{2^k(s-2)} + \cdots\right)$$Hence, $m^{2^k}+1|mD$. Let $q = m+1$ be the prime. Then, $q = m+1|m^{2^k}+1|p^a$ which implies $p = q$. Therefore, plugging $p = m+1$ in the initial equation gives us: $$m^{2^n-2} + m^{2^n-3}+\cdots+m+1 = m^n+(m+1)^a$$If $n \ge 3$, then we have two cases: $m \equiv 2, 6 \mod 8$. If $m \equiv 2 \pmod 8$, then: $$m^{2^n-2} + m^{2^n-3}+\cdots+m+1 \equiv 7 \pmod 8$$$$m^n+(m+1)^a \equiv 3^a \equiv 1,3 \pmod 8$$Hence, no solutions. If $m \equiv 6 \mod 8$, then: $$m^{2^n-2} + m^{2^n-3}+\cdots+m+1 \equiv 3 \pmod 8$$$$m^n+(m+1)^a \equiv 7^a \equiv 1,7 \pmod 8$$Hence, no solutions for this case. For $n=2$, all $m+1$ such that $m+1$ is a prime of the form $4k-1$ works (here $a=1$). Hence, the solutions are: $$(m, n) = (p-1, 2)$$where $p$ is a prime of the form $4k-1$.