can we say that  $(p^2+1)/2\ne p_0^2$ where $p$ is a Mersenne prime   Let $p=2^a-1>7$ be a Mersenne prime and so $a$ is an odd prime.
Can we say that $(p^2+1)/2$ is not equal to the square of a prime number?
Many thanks for your help 
BHZ

Suppose, $p^2-2p_1^2=-1.$ Substituting $p=2^a-1,$ we arrive at 
$$2^a(2^{a-1}-1)=(p_1-1)(p_1+1).$$ Observe, $(p_1-1,p_1+1)=2,$ so we must have the following options: $p_1-1=2^{a-1}k$ and $p_1+1=2l$ and $kl=2^{a-1}-1.$ This is impossible unless $k,l$ and thus $a$ are small. Indeed, if $k\ge 2,$ then $p_1\ge 2^a+1$ and $l\ge 2^{a-1}$ so $kl\ge2^a.$ If $k=1,$ then $p_1=2^{a-1}+1$ and $l=2^{a-2}+1$ and $kl=2^{a-2}+1=2^{a-1}-1.$ This implies $a=3.$ Otherwise, $p_1-1=2k$ and $p_1+1=2^{a-1}l$ and $kl=2^{a-1}-1.$ Again it is possible only if $k,l$ and $a$ are small. These case can be checked by hands.  
