No integers satisfying $x^2+2y^2 = p$

From $x^2\equiv 1\pmod{2}$. You got that $x$ has to be odd. You have two options:

$x,y$ odd. Then $x^2+2y^2\equiv 1+2=3\pmod{8}$.

$x$ odd, $y$ even. Then $x^2+2y^2\equiv 1+0=1\pmod{8}$.

And you are done.

Another approach is simply to compute $x^2+2y^2 \bmod 8$ for $x,y =0, \dots 7$. The result is $0, 1, 2, 3, 4, 6$. Therefore, if $z \equiv 5,7 \bmod 8$, then $z$ is not of the form $x^2+2y^2$. This has nothing to do with $z$ being prime or not.