Adding $1$ to both the given numbers we get:
$\dfrac{a^3b-1}{a+1} + 1$ is also a positive integer so we have:
$\dfrac{a^3b+a}{a+1}$ = $\dfrac{a(a^2b+1)}{a+1}$ is a positive integer
$\implies (a+1) \mid (a^2b+1)$ $\implies (a+1) \mid (((a+1) - 1)^2b+1)$ $\implies (a+1) \mid (b+1)$
Similarly,
$\dfrac{b^3a+1}{b-1} + 1$ is also a positive integer so we have:
$\dfrac{b^3a+b}{b-1}$ = $\dfrac{b(b^2a+1)}{b-1}$ is a positive integer
$\implies (b-1) | (b^2a+1)$ $\implies (b-1) | (((b-1) + 1)^2a+1)$ $\implies (b-1) | (a+1)$
Combining above $2$ results we get:
$(b-1) | (b+1)$
$\implies b=2,3$
$Case 1: b=2$ $\implies a+1|3 \implies a=2$ which is a valid solution.
$Case 2: b=3$ $\implies a+1|4 \implies a=1,3$ which are valid solutions.
Thus, all solutions are: $(2,2), (1,3), (3,3)$
$Kris17$