1 plus Perfect Power is not Power of 2

Theorem

The equation:

$1 a^n = 2^m$

has no solutions in the integers for $n, m > 1$.

This is an elementary special case of Catalan's Conjecture.

Aiming for a contradiction, suppose there is a solution.

Then:

$\ds a^n$

$=$

$\ds 2^m - 1$

$\ds $

$\equiv$

$\ds -1$

$\ds \pmod 4$

as $m > 1$

$a$ is immediately seen to be odd.

By Square Modulo 4, $n$ must also be odd.

Now:

$\ds 2^m$

$=$

$\ds a^n 1$

$\ds $

$=$

$\ds \paren {a 1} \sum_{k \mathop = 0}^{n - 1} \paren {-1}^k a^{n - k - 1}$

The latter sum has $n$ powers of $a$, which sums to an odd number.

The only odd divisor of $2^m$ is $1$.

However, if the sum is $1$, we have:

$a^n 1 = a 1$

giving $n = 1$, contradicting our constraint $n > 1$.

$\blacksquare$