1 plus Power of 2 is not Perfect Power except 9

Theorem
The only solution to:
 * $1 + 2^n = a^b$

is:
 * $\tuple {n, a, b} = \tuple {3, 3, 2}$

for positive integers $n, a, b$ with $b > 1$.

This is a special case of Catalan's Conjecture.

Proof
It suffices to prove the result for prime values of $b$.

For $n = 0$, it is clear that $1 + 2^0 = 2$ is not a perfect power.

For $n > 0$, $1 + 2^n$ is odd.

Hence for the equation to hold $a$ must be odd as well.

Writing $a = 2 m + 1$ we have:

Since all factors of $2^n$ are powers of $2$:
 * $\displaystyle \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^{i - 1}$ is a power of $2$.

But since each summand is non-negative:
 * $\displaystyle \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^{i - 1} \ge 2$

we must have $\displaystyle \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^{i - 1}$ is even.

We have:

Therefore we must have $b = 2$, the only even prime.

In that case:

So $m$ and $m + 1$ are powers of $2$.

The only $m$ satisfying this is $1$, giving the solutions:
 * $a = 2 m + 1 = 3$
 * $2^n = 3^2 - 1 = 8 \implies n = 3$