Example 6 Let the sequence of positive numbers $a_{0}, a_{1}, a_{2}, \cdots$ satisfy $\sqrt{a_{n} a_{n-2}}-\sqrt{a_{n-1} a_{n-2}}=2 a_{n-1}(n \geqslant 2)$, and $a_{0}=a_{1}=1$. Find the general term formula of this sequence.
Simplified Solution 1: Using the recursive formula, we get
$$
\begin{array}{l}
a_{0}=1, a_{1}=1=\left(2^{1}-1\right)^{2}, \\
a_{2}=9=\left(2^{2}-1\right)^{2}\left(2^{1}-1\right)^{2}, \\
a_{3}=21^{2}=\left(2^{3}-1\right)^{2}\left(2^{2}-1\right)^{2}\left(2^{1}-1\right)^{2} .
\end{array}
$$

Conjecture that when $n \in \mathbf{N}_{+}$, we have
$$
a_{n}=\left(2^{1}-1\right)^{2}\left(2^{2}-1\right)^{2} \cdots\left(2^{n}-1\right)^{2} \text {. }
$$

From $\sqrt{a_{n} a_{n-2}}-\sqrt{a_{n-1} a_{n-2}}=2 a_{n-1}$, we get
$$
\begin{array}{l}
a_{k+1}=\left(\sqrt{a_{k}}+\frac{2 a_{k}}{\sqrt{a_{k-1}}}\right)^{2} \\
=\left(2^{1}-1\right)^{2}\left(2^{2}-1\right)^{2} \cdots\left(2^{k}-1\right)^{2}\left(2^{k+1}-1\right)^{2} .
\end{array}
$$

Simplified Solution 2: Transforming the original equation, we get
$$
\sqrt{a_{n} a_{n-2}}-2 a_{n-1}=\sqrt{a_{n-1} a_{n-2}} \text {. }
$$

Dividing both sides by $\sqrt{a_{n-1} a_{n-2}}$, we get
$$
\sqrt{\frac{a_{n}}{a_{n-1}}}-2 \sqrt{\frac{a_{n-1}}{a_{n-2}}}=1 \text {. }
$$

Thus, $\sqrt{\frac{a_{n-1}}{a_{n-2}}}-2 \sqrt{\frac{a_{n-2}}{a_{n-3}}}=1$,
$$
\begin{array}{l}
\sqrt{\frac{a_{n-2}}{a_{n-3}}}-2 \sqrt{\frac{a_{n-3}}{a_{n-4}}}=1, \\
\cdots \cdots \\
\sqrt{\frac{a_{2}}{a_{1}}}-2 \sqrt{\frac{a_{1}}{a_{0}}}=1 .
\end{array}
$$
$$
\text { Multiplying (1) } \times 1+\text { (2) } \times 2+\text { (3) } \times 2^{2}+\cdots+\text { (n-1) }
$$
$2^{n-2}$, we get
$$
\sqrt{\frac{a_{n}}{a_{n-1}}}-2^{n-1} \sqrt{\frac{a_{1}}{a_{0}}}=1+2+2^{2}+\cdots+2^{n-2} .
$$

Since $a_{0}=a_{1}=1$, then
$$
\sqrt{\frac{a_{n}}{a_{n-1}}}=1+2+2^{2}+\cdots+2^{n-1}=2^{n}-1 \text {. }
$$

Thus, $a_{n}=\left(2^{n}-1\right)^{2} a_{n-1}$
$$
=\cdots=\prod_{k=1}^{n}\left(2^{k}-1\right)^{2} \text {. }
$$

Therefore, $a_{n}=\prod_{k=1}^{n}\left(2^{k}-1\right)^{2}$.