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

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

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$k\ge 2 \Rightarrow$ $$-1 \equiv x^2 \mod{4}$$ 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.