The question is
Do there exist integers $x,y$ so that $2y^2+3$ divides $x^2-2$?
Apparently, this problem can be solved with quadratic reciprocity. Is there any simpler way?
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Note that $2 y^2 + 3 \equiv 3 \mod 8$ if $y$ is even, $\equiv 5 \mod 8$ if $y$ is odd. Therefore $2 y^2 + 3$ must have some prime factor $\equiv 3$ or $5 \mod 8$. But $2$ is not a quadratic residue for such a prime: see e.g. here.