The given condition implies that
$$a^2 b k = b^2 + 3a$$
for some positive integer $k$. The equation $a^2 b k - 3a = b^2$ implies $a \mid b^2$ and $b \mid 3a$. Therefore, the numbers
$$\frac{b^2}{a}, \frac{3a}{b} \quad \text{and} \quad \frac{b^2 + 3a}{a^2 b} = \frac{b}{a^2} + \frac{3}{a b}$$
are integers. By multiplying the second and the third one, we conclude that $\frac{9a}{b^2}$ is an integer, meaning that $b^2 \mid 9a$. Now we have
$$b^2 m = 9a, \quad a n = b^2$$
for some positive integers $m$ and $n$. This implies $m n = 9$. Hence, $n \in \{1, 3, 9\}$, which means that $b^2 \in \{a, 3a, 9a\}$. The initial condition now implies $a b \mid n + 3$.

We have three possible values for $n$.

If $n = 1$, then $b^2 = a$, so the initial condition becomes $b^5 \mid 4b^2$, i.e. $b^3$ divides $4$. We obtain one solution $a = b = 1$.

If $n = 3$, then $b^2 = 3a$, so $\frac{b^5}{9} \mid 2b^2$, i.e. $b^3 \mid 18$. Since $b$ is a multiple of $3$, there is no solution in this case.

If $n = 9$, then $b^2 = 9a$, so $\frac{b^5}{81} \mid \frac{4}{3} b^2$, i.e. $b^3 \mid 108$. Since $b$ is a multiple of $3$, we obtain a solution $(a, b) = (1, 3)$.

The only solutions are $(1, 1)$ and $(1, 3)$.