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!
No. This is a special case of Mihăilescu's theorem / Catalan's Conjecture, which says that the only consecutive powers are 8 and 9.
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.