The question is the following :

*Question:*

Does there exist infinitely many primes of the form $(2m+1)^2-2^{2s+1}$ with $m,s\geq 1$ ?

*Why this could be true:*

Bunyakowsky conjecture would imply that the answer is *yes* for any given $s$.

[ Indeed, the polynomial $4X^2+4X-2^{2s+1}+1$ is irreducible ($\Delta = 2^{2s+5}$), and no integer $d\geq 1$ can divide all values $P(n)$ (since $P(0)$ is odd and $P(1)-P(0)=8$) ]

Unfortunately, as the Wikipedia page indicates, Bunyakowsky conjecture is not known to hold for a single polynomial of degree $\geq 2$.

My hope is that allowing $s$ to assume infinitely many values can help ...

*Why this could be hard:*

As was pointed out by Mike Benett in this MO question, the density of primes of this form among primes that are congruent to $1$ mod $8$ is asymptotically $0$. (As an example, there are $20453$ primes $p<2^{20}$ that are congruent to $1$ mod. $8$, and only $1334$ among them that have the required form, so that the density up to that bound is only $0.065$).

negativeprime values of your expression $(2m+1)^2 - 2^{2s+1}$? $\endgroup$ – Stefan Kohl Oct 14 '15 at 20:12realquadratic extensions. $\endgroup$ – few_reps Oct 14 '15 at 22:02