Example 1. When $a$ takes what values, the equation $\left(a^{2}-1\right) x^{2}-$ $6(3 a-1) x+72=0$ has two distinct positive integer roots?
Given $\because a^{2}-1 \neq 0$, we conclude $a \neq \pm 1$, hence
$x_{1}=\frac{12}{a+1}, x_{2}=\frac{6}{a-1}$.
From $x_{1}>0, x_{2}>0$ we know $a>1$.
Furthermore, since $x_{1}, x_{2}$ are positive integers, we know $a=2$ or 3.
However, when $a=3$, $x_{1}=x_{2}=3$ does not meet the requirements.
Therefore, when $a=2$, the equation has two distinct positive integer roots $x_{1}=4, x_{2}=6$.