Let $k>2$ and $0<\epsilon<1$ be fixed real numbers. Let $x$ be a large parameter. It suffices to show that there exists a prime pair $(p,q)$ such that $x<p<2x$ and $p^k<q\leq p^k(1+\epsilon)$ and $\left(\frac{p}{q}\right)=1$. Note that for primes $p\equiv 1\pmod{4}$, the last condition is equivalent to $\left(\frac{q}{p}\right)=1$. Let us now call a prime $p$ bad if $p\equiv 1\pmod{4}$ and there exists no prime $q$ such that $p^k<q\leq p^k(1+\epsilon)$ and $\left(\frac{q}{p}\right)=1$. The number of all primes $p\in(x,2x)$ with $p\equiv 1\pmod{4}$ is $\gg x/\log x$, hence it suffices to show that the number of bad primes $p\in(x,2x)$ is $\ll x/(\log x)^2$.

For a prime $p>2$, let $\chi_p$ (resp. $1_p$) denote the quadratic (resp. trivial) Dirichlet character mod $p$. If $p$ is bad, then with the usual notations we have
$$\sum_{\substack{p^k<n\leq p^k(1+\epsilon)\\\chi_p(n)=1}}\Lambda(n)\ll p^{k/2}=o(p^k),$$
because the primes do not contribute to the left hand side, hence also
$$\left(\psi(p^k(1+\epsilon),1_p)-\psi(p^k,1_p)\right)+\left(\psi(p^k(1+\epsilon),\chi_p)-\psi(p^k,\chi_p)\right)=o(p^k).$$
The first difference is $\gg p^k$ by Chebyshev's theorem, hence for a bad prime $p$ we have
$$\left|\psi(p^k(1+\epsilon),\chi_p)-\psi(p^k,\chi_p)\right|\gg p^k,$$
hence also
$$\max_{y<2^{k+1}x^k}\left|\psi(y,\chi_p)\right|\gg p^k.$$
On the other hand, by the usual proof of the Bombieri-Vinogradov theorem (see e.g. Chapter 24 in Huxley: The distribution of prime numbers) we can see that that
$$\sum_{x<p<2x}\;\max_{y<2^{k+1}x^k}\left|\psi(y,\chi_p)\right|\ll\frac{x^{k+1}}{(\log x)^2},$$
using that $x$ is large and $k>2$. As we observed above, for a bad prime $p\in(x,2x)$ the corresponding term on the left hand side is $\gg x^k$, whence the number of such primes is $\ll x/(\log x)^2$. The proof is complete.

**Added.** In fact the original conclusion holds for all $k>0$. One can derive this in a simpler way from Theorem 1 in Heath-Brown: A mean value estimate for real character sums (Acta Arith. 72 (1995), 235-275).