Unfortunately, this argument seems to break down when there are three primes $p,q,r$. What could happen that $n_p=n_q<n_r$. Then if $(n_r|p)=-1$ and $(n_r|q)=1$ I don't see the same trick working.

No, I am thinking that $p$ and $q$ are very near in size, which prohibits this obstruction.

Aha I suspected it should be simple!

@GerryMyerson Yes, the question is a bit open to interpretation, but I am just hoping to understand what, for instance sets $x^2y$ (for which each specialization gives a square) apart from $x^2+y$ for which almost no specialization gives a square.