How many integer pairs $(n,m)$ do exist such that $m^p-p^n=1$ where $p$ is prime? Let $p$ be prime. How many integer pairs $(n,m)$ exist such that
$$
m^p-p^n=1?
$$
When $p=2$, I have shown that the only integer pair that satisfies our condition is $(3,2)$ and when $p=3$ I have shown there are no integer pairs that satisfy our condition. 
I suspect that there are no integer pairs for any prime $p>2$, but my attempts at generalizing have failed. I have observed that we can rearrange our condition a bit:
$$
m^p-p^n=1 \iff p^n = m^p-1 = (m-1)(m^{p-1}+m^{p-2}+\cdots+m+1),
$$
Which implies that $m-1=p^k$ for some $1\le k <n$ and that
$$
p^n=\left(p^k+1\right)^p-1.
$$
Not sure what to do next. I have manipulated and massaged this relationship, but I do not see how to get anything useful from what I have. Any ideas or references about this?
 A: As you've written, setting $m=p^k+1$ where $k$ is a non-negative integer gives, by the binomial theorem,
$$p^n=(p^k+1)^p-1=\binom p1p^k+\binom p2p^{2k}+\binom p3p^{3k}+\cdots +\binom{p}{p-1}p^{(p-1)k}+\binom ppp^{pk}$$
Dividing the both sides by $p^{k+1}$ gives
$$p^{n-k-1}=1+\binom p2p^{k-1}+\binom p3p^{2k-1}+\cdots +p^{(p-2)k}+p^{pk-k-1}$$
Here, if $k-1\ge 1$, then the LHS is divisible by $p$ while the RHS isn't.
So, we have to have $k-1\lt 1$, i.e. $k=0,1$.


*

*Case 1 : $k=0$$$p^n=2^p-1\implies 2^p\equiv 1\pmod p$$Since $p$ has to be an odd prime with $\gcd(p,2)=1$, by Fermat's little theorem, we have $2^{p-1}\equiv 1\pmod p\implies 2^p\equiv 2\pmod p$. There are no odd primes satisfying both $2^p\equiv 1\pmod p$ and $2^p\equiv 2\pmod p$.

*Case 2 : $k=1$$$p^{n-2}=1+\frac{p(p-1)}{2}+\binom p3p+\cdots +p^{p-2}+p^{p-2}$$Here, if $p$ is an odd prime, then the LHS is divisible by $p$ while the RHS isn't. So, we have to have $p=2$ to have
$$2^{n}=(2+1)^2-1=2^3\implies n=3$$
It follows from the two cases that $\color{red}{(n,m,p)=(3,3,2)}$ is the only solution.
