If $2^k -1$ is a perfect square, do we have more than one solution?

Question

I am trying to solve the equation $2^k-1 = x^2$, I have got one solution $k = 1$. How to proceed further i.e. either show that the equation has no more solutions or has more.

Thanks

Answer 1

Hint: Reduce the equality modulo 4.

Answer 2

Like stated:
$k\ge 2 \Rightarrow$ $$ -1 \equiv x^2 \mod{4} $$ but

$-1$ is a non-quadratic residue modulo $4$ so there are no more.

But $k=0 \Rightarrow 2^k -1$ is a perfect square.