Example 2 (2004 Taiwan Mathematical Olympiad Problem) Find all pairs of positive integers $(a, b)$, satisfying
$$
\sqrt{\frac{a b}{2 b^{2}-a}}=
\frac{a+2 b}{4 b} .
$$
Solve: Squaring both sides of $\sqrt{\frac{a b}{2 b^{2}-a}}=\frac{a+2 b}{4 b}$ and simplifying, we get $a\left(a^{2}+4 a b+4 b^{2}\right)=2 b^{2}\left(a^{2}-4 a b+4 b^{2}\right)$,
which is $a(a+2 b)^{2}=2 b^{2}(a-2 b)^{2}$,
Since the left side of the above equation is even, let $a=2 t^{2}, t \in \mathbf{N}_{+}$, substituting into (1) and simplifying, we get $t^{2}\left(2 t^{2}+2 b\right)^{2}=b^{2}\left(2 t^{2}-2 b\right)^{2}$,
which is $t\left(t^{2}+b\right)=b\left|t^{2}-b\right|$,
Therefore, $t=\frac{b\left|t^{2}-b\right|}{t^{2}+b}$.
When $b<t^{2}$, from equation (2) we get $b^{2}+\left(t-t^{2}\right) b+t^{3}=0$,
Solving, we get $b=\frac{t}{2}\left(t-1 \pm \sqrt{t^{2}-6 t+1}\right)$.
Let $\sqrt{t^{2}-6 t+1}=s$, and $s \in \mathbf{N}_{+}$,
Squaring and simplifying, we get $(t+s-3)(t-s-3)=8$.
From this, we derive the unique set of positive integer solutions $\left\{\begin{array}{l}t=6, \\ s=1 .\end{array}\right.$
Thus, $\left\{\begin{array}{l}a=72, \\ b=18 .\end{array}\right.$ or $\left\{\begin{array}{l}a=72, \\ b=12 .\end{array}\right.$
When $b \geqslant t^{2}$, similarly, we get $t=0$, which is a contradiction.