Example 3 Find positive integers $x, y$, such that $x+y^{2}+(x, y)^{3}=x y(x, y)$.
【Analysis】Let $z=(x, y), x=a z, y=b z$. Then $(a, b)=1$.
The equation transforms into $a+b^{2} z+z^{2}=a b z^{2}$. Hence $z \mid a$. Let $a=c z$, i.e., $x=c z^{2}$. Then the equation becomes $c+b^{2}+z=b c z^{2} \Rightarrow c=\frac{b^{2}+z}{b z^{2}-1}$.
We now discuss the variable $z$.
(1) If $z=1$, then
$$
c=\frac{b^{2}+1}{b-1}=(b+1)+\frac{2}{b-1} \text {, }
$$

the positive integer $b$ can only be 2 or 3, easily yielding two solutions $(x, y)=(5,2)$ or $(5,3)$.
(2) If $z=2$, then $c=\frac{b^{2}+2}{4 b-1}$, i.e.,
$$
16 c=\frac{16 b^{2}+32}{4 b-1}=(4 b+1)+\frac{33}{4 b-1},
$$
$4 b-1$ can only be 3 or 11 (obviously not 33), easily yielding two solutions
$$
(x, y)=(4,2) \text { or }(4,6) \text {. }
$$
(3) If $z \geqslant 3$, then
$$
c z^{2}=\frac{b^{2} z^{2}+z^{3}}{b z^{2}-1}=b+\frac{z^{3}+b}{b z^{2}-1} \text {. }
$$

From $\frac{z^{3}+b}{b z^{2}-1} \geqslant 1$, we get
$$
b \leqslant \frac{z^{3}+1}{z^{2}-1}=\frac{z^{2}-z+1}{z-1}=z+\frac{1}{z-1},
$$

i.e., the positive integer $b \leqslant z$.
Therefore, $c=\frac{b^{2}+z}{b z^{2}-1} \leqslant \frac{z^{2}+z}{z^{2}-1}<2$, the positive integer $c$ can only be 1.
At this point, $b^{2}-b z^{2}+(z+1)=0$.
Noting that $\Delta=z^{4}-4(z+1) \in\left(\left(z^{2}-1\right)^{2}, z^{4}\right)$.
Thus, the equation has no positive integer solutions.
In summary, $(x, y)=(5,2),(5,3),(4,2),(4,6)$.