3. Let the sequence of positive numbers $a_{0}, a_{1}, \cdots, a_{n}, \cdots$ satisfy
$$
\sqrt{a_{n} a_{n-2}}-\sqrt{a_{n-1} a_{n-2}}=2 a_{n-1}(n \geqslant 2) \text {, }
$$

and $a_{0}=a_{1}=1$. Find the general term formula for the sequence $\left\{a_{n}\right\}$.
When $n \geqslant 2$,
$$
\begin{array}{l}
\sqrt{a_{n} a_{n-2}}-\sqrt{a_{n-1} a_{n-2}}=2 a_{n-1} \\
\Leftrightarrow \sqrt{a_{n} a_{n-2}}-\sqrt{a_{n-1} a_{n-2}}=2 \sqrt{a_{n-1}^{2}} .
\end{array}
$$

Thus, $\sqrt{\frac{a_{n}}{a_{n-1}}}-1=2 \sqrt{\frac{a_{n-1}}{a_{n-2}}}$. Let $b_{n}=$ $\sqrt{\frac{a_{n}}{a_{n-1}}}$, then $b_{n}=2 b_{n-1}+1$. Using the fixed point method, we can find $b_{n}$. Therefore, $a_{n}=\left\{\begin{array}{l}1, \quad n=0, \\ \left.\prod_{k=1}^{n}\left(2^{k}-1\right)^{2}, n \in \mathbf{N}_{+} .\right)\end{array}\right.$