<s>[INST] Let $m,n \in \mathbb{Z}^+$ and $p$ be an odd prime number. Show that $n=1$ if $2^m = p^n +1$. I could show the case that $m$ is even as follows, but not show the other case. Please give me hints!!!
When $m$ is even, put $m=2k$.
$2^m-1=(2^k-1)(2^k+1)$
By Euclidean algorithm, $2^k-1$ and $2^k+1$ are coprime.
So, these are integers to n-th power.
Put
$2^k-1=a^n, 
2^k+1=b^n$ ($1≦a<b,a$ and $b$ are odd).
$2=(b-a)(b^{n-1}+...+a^{n-1})≧2n$
∴$n=1$
 [/INST] From $p^n + 1 \ge p+1 \ge 4$ we have $m \ge 2$.
Hence $p^n = 2^m - 1 \equiv -1\pmod 4$, and thus $p^n$ cannot be a square.
This forces $n$ to be odd.
From $p^n + 1 = (p+1)(p^{n-1} - p^{n-2} + \dots + 1) = 2^m$, we see that
$$(p^{n-1} - p^{n-2} + \dots + 1) \mid 2^m$$
Since $p$ is odd, this factor, which is a sum of $n$ odd numbers, is odd as well.
This forces $p^{n-1} - p^{n-2} + \dots + 1 = 1$, and thus $p^n+1 = (p+1)(1)$.
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