Can we find positive integers $a$ and $k \geq 2$ with $2^n - 1 = a^k$? I would appreciate if somebody could help me with the following problem: 

For a given positive integer $n$, can we find positive integers $a$ and $k$ ($k\geq 2$) such that $2^n-1=a^k$? 

The answer would be no, I  guess.
Please give me some ideas!
 A: No. This is a special case of Mihăilescu's theorem / Catalan's Conjecture, which says that the only consecutive powers are 8 and 9.
A: Let $n\geq2$. (If $n<2$, we can manually find all solutions.)
Consider the equation $2^n-1=a^k$ modulo $4$. We see that both sides are congruent to $-1$, which implies $k$ is odd, because $-1$ is not a quadratic residue modulo $4$.
We can rewrite the equation $2^n-1=a^k$ as
$$2^n=(a+1)(1-a+a^2-a^3+\cdots+a^{k-1}).$$
But, as $a$ is odd, $1-a+a^2-a^3+\cdots+a^{k-1}$ is odd too. Because this factor has to be a power of $2$, it should be $1$. So $a^k+1=a+1$, which clearly is impossible if $a>1$.
So we've found that the only solution can be $a=1$. This is obviously a solution to the equation $2^n-1=a^k$.

Another proof that doesn't require Catalan's conjecture uses Zsigmondy's theorem. (An elementary proof of this less-known theorem can be found here, for example.)
From Zsigmondy's theorem we have that, if $a>2$, $a^k+1$ has at least one prime divisor that does not divide $a^m+1$ for all $m<k$. So certainly this means $a^k+1$ can never be a power of $2$.
A: Proof by counter example.
Let's pick n=2.
Then $2^2-1=3$.
I'm afraid that no integer $k \ge 2$ will fit.
