Proving a power Diophantine equation has no solutions

Question

Show that the Diophantine equation $2^n-x^m = 1$ with $x,n,m > 1$ has no solutions.

How do I show that $x^m$ can never be $1$ less than a power of $2$? I tried factoring it but it doesn't seem like I can.

Answer 1

Hint: It is clear that $x$ must be odd. Thus, if $m$ is even, $x^{m}$ is of the form $8k+1$, and therefore the highest power of $2$ that divides $x^m+1$ is $2^1$. So $m$ must odd.

Now factoring works. We have $x^m+1=(x+1)(x^{m-1}-x^{m-2}+x^{m-3}-\cdots +1)$. Note that the second term on the right has an odd number of odd terms. There are only a couple of steps left to the end.

Remark: Catalan long ago conjectured that if $x,y,m,n$ are integers $\ge 2$, the equation $x^m-y^n=1$ only holds when $x=3$, $m=2$, $y=2$, and $n=3$. This conjecture was proved not many years ago by Mihailescu.