Find all primes $p$ such that $(2^{p-1}-1)/p$ is a perfect square
Hint: Let $p=2k+1$ where $k \in \mathbb{N},$ then $2^{2k}-1=(2^k-1)(2^k+1)=pm^2.$ We know that $\gcd(2^k-1,2^k+1)=1$ since they are consecutive odd integers, so the equation breaks into $2^k-1=px^2, 2^k+1=y^2$ or $2^k-1=x^2, 2^k+1=py^2.$
Easy investigation shows that the only solutions are $p=3,7.$ I will leave it to you to fill the gaps. You may also interested in this.