Answer: $q = 7$, $p = 19$.

Note that $q \le p$, otherwise LHS $< 0$, RHS $> 0$. We have
$$\begin{aligned} p^3 - q^3 &= pq^3 - 1 \Leftrightarrow p^3 + 1 = pq^3 + q^3 \\ &\Leftrightarrow (p+1)(p^2-p+1) = q^3(p+1) \\ &\Leftrightarrow p(p-1) = q^3 - 1 \\ &\Leftrightarrow p(p-1) = (q-1)(q^2+q+1). \end{aligned}$$
Since $p(p-1) \nmid p$, we have $(q-1)(q^2+q+1) \nmid p$. But $q-1 \nmid p$ since $q-1 < p$, so $q^2+q+1 \nmid p$, i.e., $q^2+q+1 = kp$, $k \in \mathbb{N}$. Therefore
$$p(p-1) = (q-1)kp \Leftrightarrow p-1 = k(q-1) \Leftrightarrow p = k(q-1)+1.$$
Then
$$q^2 + q + 1 = kp \Leftrightarrow q^2 + q + 1 = k^2(q - 1) + k. \quad (1)$$
If $k > 3$ we have
$$\begin{aligned} q^2 + q + 1 = k^2(q - 1) + k &\implies (q - 1)(q + 2) + 3 = k^2(q - 1) + k \\ &\implies (q - 1)(q + 2) - k^2(q - 1) = k - 3. \end{aligned}$$
Hence,
$$\begin{gathered} (k-3) \mid (q-1) \implies k \ge q-1+3=q+2 \implies \\ (q-1)(q+2) = k^2(q-1) + k - 3 \ge (q+2)^2(q-1) + q - 1 > (q-1)(q+2), \\ \text{which is impossible. So, we have no solutions when } k > 3. \text{ It remains to} \\ \text{consider three cases: } k=1,2,3. \end{gathered}$$
If $k=1$ from (1) we have $q^2 + q + 1 = q$. Hence $q^2 + 1 = 0$, which is impossible.

If $k=2$ we have $q^2 + q + 1 = 4(q - 1) + 2$, which gives $q^2 - 3q + 3 = 0$. It is easy to see that this equation has no real solution.

If $k=3$ from (1) we have
$$q^2 + q + 1 = 9(q - 1) + 3 \Leftrightarrow q^2 - 8q + 7 = 0.$$
It is easy to see that either $q=1$ or $q=7$. But 1 is not prime, so $q=7$, then $p=3(q-1)+1=19$.