If $a\equiv1\pmod{4}$, let $a=4g+1$ and take $x=g<a$ $\implies 4x^2+a=4g^2+4g+1=(2g+1 )^2$, Absurd! $\implies a\equiv-1\pmod{4}, \boxed{a=4b-1}$ Assertion: $b=2^c$ Proof: If $p$ is an odd prime divisor of $b$, with $b=py$, take $x=\dfrac{p-1}{2}$ $\implies 4x^2+a=(p-1)^ 2+4py-1=p^2-2p+4py$. Absurd, because $p^2-2p+4py\neq{p}$ and will never be prime, therefore, $b$ does not have any odd prime factor. Proven! Thus, $4x^2+a=4x^2+2^{c+2}-1$, where we find the solution $a=3.7$ when $c=0.1$. Let's prove that there is no solution other than these. If $c\geq2$, take $x\equiv0\pmod{7} \implies 2^{c+2}\not\equiv1\pmod{7} \implies c\equiv0,-1\pmod{3}$. Now take $x\equiv1\pmod{7}$ to find $c\not\equiv0\pmod{7}$ Finally, take $x\equiv2\pmod{13}$ to find $2^{c+2}\not\equiv-2\pmod{13} \implies c+2\not\equiv1\pmod{12} \implies c\not\equiv-1\pmod{3}$. Absurd, because in this way, $c$ does not exist $ \implies 1\geq{c}\geq0$, and so the only possible integers $a$ are $\boxed{3,7}$.