The only pair of primes $(p,q)$ satisfying the Diophantine equation $(1) \;\; p^3 - q^3 = pq^3 - 1$ is $(p,q)=(19,7)$. Proof: Equation (1) is equivalent to $(p + 1)(p^2 - p + 1) = (p + 1)q^3$, i.e. (since $p \geq 2$) $(2) \;\; p^2 - p + 1 = q^3$, or equivalently $(3) \;\; p(p - 1) = (q - 1)(q^2 + q + 1)$. Notice that if $p \mid q-1$, then $p<q$, yielding $pq^3<1$, which obviously is impossible. Hence $p \nmid q-1$, which combined with equation (3) result in $(4) \;\; p\mid q^2 + q + 1$. Combining equation (3) and condition (4) we obtain $q - 1 \mid p - 1$, i.e. there is a positive integer $t$ s.t. $(5) \;\; p = (q - 1)t + 1$. Combining equations (3) and (5) we find that $[(q - 1)t + 1]t = q^2 + q + 1$, i.e. $(6) \;\; q^2 - (t^2 - 1)q + t^2 - t + 1 = 0$. The discriminant $d$ of equation (6) is given by $d^2 = (t^2 - 1)^2 - 4(t^2 - t + 1) = t^4 - 6t^2 + 4t - 4$, yielding $t \geq 3$. Moreover $(t^2 - 3)^2 < t^4 - 6t^2 + 4t - 4 < (t^2 - 2)^2$ i.e. $t^2 - 3 < d < t^2 - 2$ when $t \geq 4$. Consequently $t=3$, which inserted in equation (6) result in $q^2 - 8q + 7 = 0$, which means $(q - 1)(q - 7) = 0$, yielding (since $q$ is a prime) $q=7$. Hence by formula (6) we obtain $p = (7 - 1) \cdot 3 + 1 = 6 \cdot 3 + 1 = 18 + 1 = 19$. Summa summarum, the only solution in primes of equation (1) is $(p,q)=(19,7)$. q.e.d.

Here is my solution: https://calimath.org/pdf/BelarusTST2016-3.pdf And I uploaded the solution with motivation to: https://youtu.be/3oXGUU4Xa8A