Example 5. For what value of $m$ does the equation $\left(m^{2}-1\right) x^{2}$ $-6(3 m-1) x+72=0$ have positive integer solutions?
Solve $\because \Delta=[6(m-3)]^{2}$,
$$
\begin{aligned}
\therefore \quad x & =\frac{6(3 m-1) \pm 6(m-3)}{2\left(m^{2}-1\right)} \\
& =\frac{2[(3 m-1) \pm(m-3)]}{m^{2}-1} .
\end{aligned}
$$

Thus, $x_{1}=\frac{12}{m+1}, \quad x_{2}=\frac{6}{m-1}$.
$\because x_{1}>0, x_{2}>0, m$ is an integer,
$\therefore$ For $x_{1}$, $m+1$ must be a divisor of 12.
Let $m+1=1,2,3,4,6,12$,
Then $m=0,1,2,3,5,11$.
$\because$ For $x_{2}$, $m-1$ must be a divisor of 6,
$$
\therefore m-1=1,2,3,6 \text {. }
$$

Thus, $m=2,3,4,7$.
According to the problem, $m$ must make $x_{1}, x_{2}$ both integers, so take $m=2,3$.
Therefore, when $m=2,3$, the equation has positive integer solutions.