Solve: Let $b^{2}+3 a=a^{2} b k\left(k \in \mathbf{Z}_{+}\right)$, $(a, b)=d$.
Set $a=a_{1} d, b=b_{1} d,\left(a_{1}, b_{1}\right)=1$. Then $b_{1}^{2} d+3 a_{1}=a_{1}^{2} b_{1} d^{2} k$
$\Rightarrow d \mid 3 a_{1}$ and $a_{1} \mid d$.
Set $3 a_{1}=d m, d=n a_{1}\left(m, n \in \mathbf{Z}_{+}\right)$.
Then $m n=3$.
(1) When $m=1, n=3$,
$d=3 a_{1}, b_{1}^{2}+1=3 a_{1}^{3} b_{1} k$.
Thus $b_{1} \mid 1, b_{1}=1$, at this time, $a_{1}$ has no solution.
(2) When $m=3, n=1$,
$$
d=a_{1}, b_{1}^{2}+3=a_{1}^{3} b_{1} k \text {. }
$$

Thus $b_{1} \mid 3, b_{1}=1$ or 3.
(i) If $b_{1}=1$, then $a_{1}^{3} k=4$, only $a_{1}=1, k=4$, at this time, $a=b=1$, which meets the requirement.
(ii) If $b_{1}=3$, still $a_{1}^{3} k=4, a_{1}=1, k=4$, at this time, $a=1, b=3$, which meets the requirement.

In summary, the required positive integer pairs $(a, b)=(1,1)$, $(1,3)$.
Find all pairs of positive integers $(a, b)$ such that $a^{2} b \mid\left(b^{2}+3 a\right)$ $\square$