Example 4 Given the sequence $\left\{a_{n}\right\}$, where $a_{1}=1, a_{2}=5$, and satisfies
$$
a_{n+1}=\frac{a_{n} a_{n-1}}{\sqrt{a_{n}^{2}+a_{n-1}^{2}+1}} .
$$

Find $a_{n}$.
Analysis: This recursive sequence is a non-linear fractional type with a more complex denominator than the numerator. We can first square and take the reciprocal, then factorize and substitute, and finally take the logarithm to reduce it to a homogeneous linear recursive sequence.
Solution: Squaring equation (1) and taking the reciprocal, we get
$$
\begin{array}{l}
\frac{1}{a_{n+1}^{2}}=\frac{1}{a_{n}^{2} a_{n-1}^{2}}+\frac{1}{a_{n}^{2}}+\frac{1}{a_{n-1}^{2}} \\
=\left(\frac{1}{a_{n}^{2}}+1\right)\left(\frac{1}{a_{n-1}^{2}}+1\right)-1 . \\
\text { Thus, } \frac{1}{a_{n+1}^{2}}+1=\left(\frac{1}{a_{n}^{2}}+1\right)\left(\frac{1}{a_{n-1}^{2}}+1\right) . \\
\text { Let } b_{n}=\frac{1}{a_{n}^{2}}+1, \text { then } \\
b_{n+1}=b_{n} b_{n-1}, b_{1}=2, b_{2}=\frac{26}{25} .
\end{array}
$$

It is clear that $b_{n}>0$.
Taking the common logarithm of both sides of $b_{n+1}=b_{n} b_{n-1}$, we get
$\lg b_{n+1}=\lg b_{n}+\lg b_{n-1}$.
Let $F_{n}=\lg b_{n}$, then,
$$
F_{n+1}=F_{n}+F_{n-1}, F_{1}=\lg 2, F_{2}=\lg \frac{26}{25},
$$

This is the Fibonacci sequence, whose characteristic equation is $x^{2}-x-1=0$.
Solving, we get $x_{1}=\frac{1+\sqrt{5}}{2}, x_{2}=\frac{1-\sqrt{5}}{2}$.
Assume $F_{n}=A\left(\frac{1+\sqrt{5}}{2}\right)^{n}+B\left(\frac{1-\sqrt{5}}{2}\right)^{n}$.
Given $F_{1}=\lg 2, F_{2}=\lg \frac{26}{25}$, we can further determine $F_{n}$.
Thus, $b_{n}=10^{F_{n}}$, i.e., $\frac{1}{a_{n}^{2}}+1=10^{F_{n}}$.
Therefore, $a_{n}=\sqrt{\frac{1}{10^{F_{n}}-1}}$.