Example $\mathbf{4}$ Find all pairs of positive integers $(m, n)$, satisfying $\left(2 m^{2}+n^{2}\right) \mid(3 m n+3 m)$.
(2012, Slovenia National Team Selection Exam)
From the problem, we set
$3 m n+3 m=k\left(2 m^{2}+n^{2}\right)(k \in \mathbf{Z})$.
Then $3 m(n+1)=k\left(2 m^{2}+n^{2}\right) \geqslant 2 \sqrt{2} k m$.
Thus, $k \leqslant \frac{3(n+1)}{2 \sqrt{2} n}=\frac{3}{2 \sqrt{2}}\left(1+\frac{1}{n}\right) \leqslant \frac{3}{\sqrt{2}}$.
From this, we get $k \leqslant 2$.
Considering $3 m n+3 m=k\left(2 m^{2}+n^{2}\right)$ as a quadratic equation in $n$,
$k n^{2}-3 m n+2 k m-3 m=0$.
Thus, $\Delta=9 m^{2}-8 k^{2} m^{2}+12 k m$.
If $k=2$, then $\Delta=24 m-23 m^{2}$.
By $\Delta \geqslant 0$, we get $m=1$.
From equation (1), we get $n=1$.
If $k=1$, then the discriminant is a non-negative integer $t$ squared, i.e.,
$$
\begin{array}{l}
t^{2}=\Delta=m^{2}+12 m=(m+6)^{2}-36 \\
\Rightarrow 36=(m+6+t)(m+6-t) .
\end{array}
$$

Since $t \geqslant 0, m>0$, we have
$$
m+6+t>0 \text {. }
$$

Thus, $m+6-t>0$.
Clearly, $m+6+t$ and $m+6-t$ have the same parity.
Hence, $m+6+t=m+6-t=6$

or $m+6+t=18, m+6-t=2$.
In the second case, we get
$$
m=4, t=8 \text {. }
$$

Therefore, equation (1) has solutions $n=2,10$.
In summary, the positive integer pairs $(m, n)$ that satisfy the problem's requirements are $(1,1),(4,2),(4,10)$.