Example 2 Given $x_{1}=1, x_{n+1}=\sqrt{2 x_{n}^{4}+6 x_{n}^{2}+3}\left(n \geqslant 1, n \in \mathbf{N}^{*}\right)$, find $x_{n}$.
Solution: The recurrence relation of $\left\{x_{n}\right\}$ can be transformed into $x_{n+1}^{2}=2 x_{n}^{4}+6 x_{n}^{2}+3$.
Let $y_{n}=x_{n}^{2}$, then $y_{1}=1, y_{n}=2 y_{n}^{2}+6 y_{n}+3=f\left(x_{n}\right)$, where $f(x)=2 x^{2}+6 x+3$.
Solving the equation $f(x)=x$, i.e., $2 x^{3}+5 x+3=0$, we get two fixed points of $f(x)$: $\alpha=-\frac{3}{2}, \beta=-1$.
Thus, $y_{n+1}-\left(-\frac{3}{2}\right)=2 y_{n}^{2}+6 y_{n}+\frac{9}{2}=2\left[y_{n}-\left(-\frac{3}{2}\right)\right]^{2}$.
Then, $y_{n}+\frac{3}{2}=2\left(y_{n-1}+\frac{3}{2}\right)^{2}=2 \cdot\left[2\left(y_{n-2}+\frac{3}{2}\right)^{2}\right]^{2}=2^{1+2}\left(y_{n-2}+\frac{3}{2}\right) 2^{2}$
$$
=\cdots=2^{1+2+\cdots-2^{n+2}}\left(y_{1}+\frac{3}{2}\right) 2^{n-1}=2^{2^{n}-1}\left(y_{1}+\frac{3}{2}\right)^{2^{n-1}}=\frac{1}{2} \cdot 5^{2^{n-1}} \text {. }
$$

It can be easily proven by mathematical induction that $x_{n}>0$, so $x_{n}=\sqrt{y_{n}}=\sqrt{\frac{1}{2}\left(5^{2^{n}}-3\right)}\left(n \in \mathbf{N}^{*}\right)$.
Analysis: Generally, it can be proven that if $\alpha=-\frac{b}{2 a}$ is exactly a fixed point of $f(x)=a x^{2}+b x+c(a \neq 0)$, then the recurrence relation $x_{n+1}=a x_{n}^{2}+b x_{n}+c$ can be transformed into $x_{n+1}-\alpha=a\left(x_{n}-\alpha\right)^{2}$.