Example 11 Try to find all positive integers $a$ such that the equation $a x^{2}+2(2 a-1) x+4(a-3)=0$ has at least one integer solution.
Solution: Since the parameter $a$ is linear, we can express $a$ in terms of $x$, i.e.,
$$
a=\frac{2(x+6)}{(x+2)^{2}}.
$$

Since $a$ is a positive integer, then $\frac{2(x+6)}{(x+2)^{2}} \geqslant 1$.
Solving this, we get $-4 \leqslant x \leqslant 2$ and $x \neq -2$.
Thus, $x=-4,-3,-1,0,1,2$.
Substituting these into (1) yields $a=1,3,6,10$.