The answer is $\boxed{p=3,7}$.

If $p=2$, it is not true. now we let $p\geq 3$. $\frac{2^{p+1}-4}{p}=\frac{4(2^{\frac {p-1}2}-1)(2^{\frac {p-1}2}+1)}{p}\Rightarrow 2^{\frac {p-1}2}-1$ or $2^{\frac {p-1}2}+1$ is a perfect square. If $2^{\frac {p-1}2}-1$ is a perfect square, it is obviously true when $p=3$. Now let $p\geq 5$, then $2^{\frac {p-1}2}-1\equiv 3\pmod 4$ is not a perfect square, contradiction! If $2^{\frac {p-1}2}+1$ is a perfect square, then $p\geq 7$. Let $2^{\frac {p-1}2}+1=k^2$. Then $2^{\frac {p-1}2}=k^2-1=(k+1)(k-1)$. Notice that $k\equiv k^2\equiv 1\pmod 2$, $\gcd (k+1,k-1)=2$. Therefore $\gcd\left(\frac{k-1}2,\frac{k+1}2\right) =1$. As ${2^{\frac{p-5}2}}=\frac{k-1}2\cdot\frac{k+1}2$, $\frac{k-1}2=1\Rightarrow k=3\Rightarrow p=7$. $\therefore p=3$ or $7.\blacksquare$

do $(2^{\frac{p-1}{2}}-1)(2^{\frac{p-1}{2}}+1)=pa^2$ and then split in cases. Solve both cases by Catalan.