<s>[INST] When is $2^n \pm 1$ a perfect power Is there an easy way of showing that $2^n \pm 1$ is never a perfect power, except for $2^3 + 1 = 3^2 $?
I know that  Catalan's conjecture (or Mihăilescu's theorem) gives the result directly, but I'm hopefully for a more elementary method.

I can show that it is never a square, except for $2^3 + 1$.
Proof: Cases $n=1, 2, 3$ are easily dealt with. Henceforth, let $n\geq4$.
$2^n -1 \equiv 3 \pmod{4}$ hence is never a square.   
If $2^n +1 =x^2$, then $2^n = (x-1)(x+1)$ and both of these are powers of 2. Thus we must have $(x-1) = 2, (x+1) = 4$. This gives the solution of $2^3 + 1 = 3^2$.

Let's now do an odd prime.
Say $2^n +1 = x^p$. Then $2^n = x^p - 1 = (x-1)(x^{p-1}+x^{p-2} + \ldots +1)$, so both terms are powers of 2. We have $ x = 2^m+1$ is odd. But the other term is the sum of $p$ odd numbers, hence is odd. Since this is clearly greater than 1, we get a contradiction.
Say $2^n -1 = x^p$. Then $2^n = x^p + 1 = (x+1)(x^{p-1} - x^{p-2} + \ldots -x +1 )$, so both terms are powers of 2. We have $x = 2^m -1$ is odd. But the other term is the sum of $p$ odd numbers, hence is odd. Since $x^p + 1 \neq x+1$ except for $p=1$, this means that the term is greater than 1. Hence we get a contradiction.
 [/INST] (This proof was completed with an insightful comment from Gottfried. I'm placing it as an answer so that it is readily seen that an answer exists, as opposed to just leaving it in the question.)
Suppose we have $ 2^n \pm 1 = x^k$ for some positive integers $x, k$. Cases $n=1, 2, 3$ are easily dealt with. $n=3$ yields the solution $2^3 + 1 = 3^2$. Henceforth, let $n\geq4$. Furthermore, a simple check shows that $x\neq 1, 2$.
First, we will deal with the power $k=2l$ being even. Rewriting $x^{2l}$ as $(x^l)^2$, we may assume that $k=2$.
Proof:  $2^n -1 \equiv 3 \pmod{4}$ hence is never a square.   
If $2^n +1 =x^2$, then $2^n = (x-1)(x+1)$ and both of these are powers of 2 that differ by 2. Thus we must have $(x-1) = 2, (x+1) = 4$, which gives $2^n = 8$ so $n=3$ (reject). $_\square$
Now, we deal with $k$ odd.
Proof: Say $2^n +1 = x^k$. Then $2^n = x^k - 1 = (x-1)(x^{k-1}+x^{k-2} + \ldots +1)$, so both terms are powers of 2. We have $ x -1$ is a power of 2 greater than 1, hence is even. Thus $x$ is odd. But the other term is the sum of $k$ odd numbers, hence is odd. Since this is clearly greater than 1, we get a contradiction.
Say $2^n -1 = x^k$. Then $2^n = x^k + 1 = (x+1)(x^{k-1} - x^{k-2} + \ldots -x +1 )$, so both terms are powers of 2. We have $x+1$ is a power of 2 that is greater than 1, hence is even. Thus $x$ is odd. But the other term is the sum of $p$ odd numbers, hence is odd. Since $x^p + 1 \neq x+1$ except for $p=1$, this means that the term is greater than 1. Hence we get a contradiction. $ _\square$
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