Let $p=2^a-1$ be a Mersenne prime and so $a$ is an odd prime if $p>7$. We know that if $p=7$, then $(p^2+1)/2$ is equal to $5^2$.
Can we prove that if $p>7$ , then $(p^2+1)/2$ is not equal to the powers of a prime number $t$, i.e. there exists no prime number $t$ such that $(p^2+1)/2=t^\beta$ for some $\beta>0$?


One more question: 


Can we prove that if there exists a prime number $q$ such that $q$ divides $(p^2+1)/2$, then $q^2$ does not didivde $(p^2+1)/2$?


Thanks for your help
BHZ



Factorization of $p^2+1$ where $p$ is a Mersenne prime  
If $p=2^a-1$ and $(p^2+1)/2=t^b$, then $2^{2a-1}-2^a+1=t^b$. If $a \equiv 1 \pmod 4$ and $b$ is even then we get an integral point on the curve $y^2=2x^8-2x^4+1$, so there are only finitely many such $p$. Likewise, we can deal with $a \equiv 3 \pmod 4$, using another curve. For $b>1$ odd, you get a curve that depends on $t$, so it's not immediate, but something similar might work. Note that I didn't use that $p$ was prime, except to assume that $a$ is odd.


Now, I don't see an obstruction right away in the case $b=1$, in particular there should be infinitely many primes $t=2^{2a-1}-2^a+1$. With the additional requirement that $p$ is prime, things get a little muddier.


Finally, for the last question, it should follow from the ABC conjecture. 


