Fermat Squares

I don't know if there's an easier way to do the problem, but here was my approach.

Write $2^{p-1}-1= px^2$ . Then $\left( 2^{\frac{p-1}2} - 1 \right) \left( 2^{\frac{p-1}2} + 1 \right) = px^2.$ These two factors are relatively prime. So by unique factorization we get that one of them is a square and the other is $p$ times a square.

There are two cases.

If $2^{\frac{p-1}2} - 1$ is a square, then considering the equation mod $4$ shows that $2^{\frac{p-1}2}$ is $1$ or $2$ mod $4$ , so the only possibility to consider is $p = 3$ , which happens to work.

If $2^{\frac{p-1}2} + 1 = y^2$ , then we can factor $2^{\frac{p-1}2} = (y-1)(y+1)$ . Since the two factors are powers of $2$ that differ by $2$ , they can only be $2$ and $4$ , so $y = 3$ and $p = 7$ , which also happens to work. So there are $\fbox{2}$ values in all.