Example 5 (1993 National High School League Question) Let the sequence of positive integers $a_{0}, a_{1}, a_{2}, \cdots$ satisfy $a_{0}=a_{1}=1$ and $\sqrt{a_{n} \cdot a_{n-2}}-\sqrt{a_{n-1} \cdot a_{n-2}}=2 a_{n-1}(n=2,3, \cdots)$, find the general term formula of this sequence.
Solving: By dividing both sides of the original recurrence relation by $\sqrt{a_{n-1} a_{n-2}}$, we get
$$
\sqrt{\frac{a_{n}}{a_{n-1}}}-1=2 \sqrt{\frac{a_{n-1}}{a_{n-2}}}(n \geqslant 2) .
$$

Let $b_{n}=\sqrt{\frac{a_{n}}{a_{n-1}}}$, then $b_{1}=1$, and $b_{n}-1=2 b_{n-1}(n \geqslant 2)$, which means $b_{n}+1=2\left(b_{n-1}+1\right)$, hence $b_{n}+1$ is a geometric sequence with the first term $b_{1}+1=2$ and common ratio $q=2$, so
$$
b_{n+1}+1=2 \cdot 2^{n-1}=2^{n}, b_{n}=2^{n}-1 \text {, }
$$

which means $\frac{a_{n}}{a_{n-1}}=b_{n}^{2}=\left(2^{n}-1\right)^{2}$, thus
$$
a_{n}=\frac{a_{n}}{a_{n-1}} \cdot \frac{a_{n-1}}{a_{n-2}} \cdot \cdots \cdot \frac{a_{1}}{a_{0}} \cdot a_{0}=\left(2^{n}-1\right)^{2}\left(2^{n-1}-1\right)^{2} \cdots\left(2^{1}-1\right)^{2} .
$$