TITLE: Why cannot $2^x=m^n+1$?
QUESTION [8 upvotes]: $x,n,m >1$ and $x,n,m \in \mathbb{Z}$
I've tried to solve it myself, but I'm getting nowhere, so apologies if it's an irritatingly basic question.
Whilst I'm on it, is it true that $y^x \ne m^n+1$ ($y>1, y \in \mathbb{Z}$)? There's probably a glaringly obvious counterexample I'm overlooking.

REPLY [9 votes]: The Diophantine equation is $$2^x = 1 + m^n.$$

Since the left hand side is even $m$ must be odd, let $m = 2r+1$.
Since $m^n + 1$ factors into $(m+1)(\cdots)$ when $n$ is odd, $n$ must be even, let $n = 2u$.

Putting these two facts together and applying the binomial theorem gives $$m^n = (2r+1)^{2u} = (4r(r+1)+1)^u = \sum_{k=0}^u \binom{u}{k} [4r(r+1)]^k$$
Plugging this into our original equation, and pulling out the first term of the sum we have
$$\begin{array}{rcl}
2^x
&=& 1 + \binom{u}{0} [4r(r+1)]^0 + 4 \sum_{k=1}^u \binom{u}{k} 4^{k-1}[r(r+1)]^k \\
&=& 2 + 4 M \\
\end{array}$$ for some $M$, this is impossible unless $M=0$, then $x=1$ and $u=0$ and $m$ can be anything.