Prime Suspects of Perfect Squares

First we notice that for $p=2$ the expression $2^{p-1}-1$ is not a square.Hence we conclude that $p$ is odd prime.

According to Fermat's Little Theorem $p$ divides $2^{p-1}-1$ .

For any integer $a$ , $2^{p-1}-1=pa^2$

As $p$ is a odd prime ,we can say $(2^{(p-1)/2}+1)c=pa^2$

As both the factors of $pa^2$ are consecutive odd integer $p$ divides anyone of them.

Thus case 1 $(2^{(p-1)/2}+1)=x^2$ & $2^{(p-1)/2}-1)=py^2$

As $(2^{(p-1)/2}+1)=x^2$ we can conclude $(2^{(p-1)/2}=(x+1)(x-1)$

the only solution from this is $p=7$ & $x=3$

case 2 $(2^{(p-1)/2}+1)=py^2$ & $2^{(p-1)/2}-1)=x^2$

If $p>3$ $x^2$ leaves a remainder of $3$ when divided by $4$ .Hence we get that the only value is $p=3$

So the sum of the values of $p$ is $7+3=\boxed{10}$