 Solution

Because $p$ is odd, $x$ must be odd. Every odd square is $1$ modulo eight. If $y$ is even, $2y^2$ is zero modulo eight; if $y$ is odd, it is two modulo eight. ThusMore precisely, the first case has even $y$ and the second has odd $y$. These are necessary conditions on the congruence class, as requested; the elementary calculation alone does not prove their sufficiency.

Equivalently, reduction modulo $p$ makes $-2$ a quadratic residue: $p$ cannot divide $y$, since that would also force $p\mid x$ and hence $p^2\mid p$. The two supplementary laws and multiplicativity of the Legendre symbol give the same two allowed classes.