Reasonable exceptions allowed. Example: $n=2$.

Suppose $q$ is odd. Let $p$ be so $pq\equiv -1$ (mod 8). Then $q\neq$ 2nd power (mod $p$) is the same as $\left(\frac{q}{p}\right)=-1$. We can obtain this from quadratic reciprocity law: $$\left(\frac{q}{p}\right)\left(\frac{p}{q}\right)\equiv(-1)^{\frac{p-1}{2}\frac{q-1}{2}}=(-1)^{\frac{pq+1}{4}}(-1)^{\frac{p+q}{2}}=(-1)^{\frac{p+q}{2}}$$ where the last equality holds since $pq\equiv -1$ (mod 8) $\Rightarrow$ $\frac{pq+1}{4}\equiv 0$ (mod 2).

Then the condition $\left(\frac{q}{p}\right)=-1$ is equivalent to $\left(\frac{p}{q}\right)=-(-1)^{\frac{p+q}{2}} = -(-1)^{\frac{-q^{-1}+q}{2}}$, where the $q^{-1}$ is taken (mod 8). In particular, for any $p\equiv -(-1)^{\frac{q-q^{-1}}{2}}$ (mod $q$) and $pq\equiv -1$ (mod 8), we will have $q\neq$ 2nd power (mod $p$).

I was trying to repeat this argument for higher powers. I can find conditions so that, say, $n$-th residue symbol will equal to 1 (for certain $n$ anyways). However, that does not guarantee that $q$ will not be a $n$-th power (mod $p$).

Any thoughts appreciated.