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

(Solution of E. Dauhiala, B. Gilevich, A. Zhuk, A. Semchankau.) We have
$$q^3 = p^2 - p + 1 \Leftrightarrow (q-1)(q^2+q+1) = p(p-1). \quad (1)$$
If $(q-1) \nmid p$, then $q \ge p+1$, so $q^3 > p^2 - p + 1$.
Hence, $(q^2+q+1) \nmid p$, i.e.
$$q^2 + q + 1 = k p \quad (2)$$
for some $k \in \mathbb{N}$. It follows that $k(q-1) = p-1$, or $p = kq - k + 1$. Substituting this expression in (2) we obtain $q^2 + (1-k^2)q + (k^2-k+1) = 0$. The discriminant of this quadratic on $q$ equation
$$D = (k^2 - 1)^2 - (4k^2 - k + 1) = k^4 - 6k^2 + 4k - 3$$
must be a perfect square. But it is easy to check that
$$(k^2 - 3)^2 < k^4 - 6k^2 + 4k - 3 < (k^2 - 1)^2 \quad \text{for any } k > 3.$$
Moreover, the equation $k^4 - 6k^2 + 4k - 3 = (k^2 - 2)^2$ has no solutions.
Hence, $k \le 3$. For $k = 1$ we have $D = -4$, and for $k = 2$ we have $D = -3$, which is impossible. For $k = 3$ we obtain $D = 6^2$, so $q = (8 \pm 6)/2$. Thus $q = 7$ and $p = k(q-1) + 1 = 19$. It remains to note the primes $p = 19, q = 7$ do satisfy the equation.

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

We have
$$q^3 = p^2 - p + 1 \Leftrightarrow (q-1)(q^2 + q + 1) = p(p-1). \quad (1)$$
If $(q-1) \nmid p$, then $q \ge p+1$, so $q^3 > p^2 - p + 1$.
Hence, $(q^2 + q + 1) \nmid p$, i.e.
$$q^2 + q + 1 = k p \quad (2)$$
for some $k \in \mathbb{N}$. It follows that $k(q-1) = p-1$, or $p = kq - k + 1$. Substituting this expression in (2) we obtain $q^2 + (1-k^2)q + (k^2-k+1) = 0$. The discriminant of this quadratic on $q$ equation
$$D = (k^2 - 1)^2 - (4k^2 - k + 1) = k^4 - 6k^2 + 4k - 3$$
must be a perfect square. But it is easy to check that
$$(k^2 - 3)^2 < k^4 - 6k^2 + 4k - 3 < (k^2 - 1)^2 \quad \text{for any } k > 3.$$
Moreover, the equation $k^4 - 6k^2 + 4k - 3 = (k^2 - 2)^2$ has no solutions.
Hence, $k \le 3$. For $k = 1$ we have $D = -4$, and for $k = 2$ we have $D = -3$, which is impossible. For $k = 3$ we obtain $D = 6^2$, so $q = (8 \pm 6)/2$. Thus $q = 7$ and $p = k(q-1) + 1 = 19$. It remains to note the primes $p = 19$, $q = 7$ do satisfy the equation.