Let $p$ be a prime congruent to $1$ mod. 8. 


If $p= 17$ one has : $p+ 8 = 5 ^2$.


If $p= 41$ one has : $p+ 8 = 7 ^2$.


If $p= 73$ one has : $p+ 8 = 9 ^2$.


If $p= 89$ one has : $p+ 32 = 11 ^2$.


If $p= 97$ one has : $p+ 128 = 15 ^2$.


If $p= 113$ one has : $p+ 8 = 11 ^2$.


If $p= 137$ one has : $p+ 32 = 13 ^2$.


If $p= 193$ one has : $p+ 32 = 15 ^2$.


If $p= 233$ one has : $p+ 128 = 19 ^2$.


For $p=241$, there is no value $s\leq 4000$ for which $p+2^{2s+1}$ is a square. 


Question 1 : Is there a simple way to prove that no such $s$ can exist ?


Question 2 : Can one estimate the density of those primes $p\equiv 1$ mod. 8 that can be written under the form
$$p=m^2-2^{2s+1}\ \ \ ?$$



Does the equation $241+2^{2s+1}=m^2$ have a solution?
To answer your first question: there is indeed no $s$ such that
$241+2^{2s+1}$ is a perfect square. -- **Proof:** $2^{2s+1}$ is always
congruent to either $2$, $8$ or $32$ modulo $63$, which makes
$241+2^{2s+1}$ congruent to either $21$, $54$ or $60$ modulo $63$.
However none of these values is a quadratic residue modulo $63$,
and thus $241+2^{2s+1}$ cannot be a perfect square, as claimed.


