Example 7 Given $x_{1}=1, x_{n+1}=\sqrt{2 x_{n}^{4}+6 x_{n}^{2}+3}(n \geqslant 1)$, find $x_{n}$.
Solve the recurrence relation of $\left\{x_{n}\right\}$ which 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 $f(x)=2 x^{2}+6 x+3$, solve the equation $f(x)=x$, i.e., $2 x^{2}+5 x+3=0$, to get the two fixed points of $f(x)$: $\alpha=-\frac{3}{2}, \beta=-1$.
$$
y_{n+1}-\left(-\frac{3}{2}\right)=2 y_{n}^{2}+6 y_{n}+\frac{9}{2}=2\left[y_{n}-\left(-\frac{3}{2}\right)\right]^{2} .
$$

Thus, $y_{n}+\frac{3}{2}=2\left(y_{n-1}+\frac{3}{2}\right)^{2}=2 \cdot\left[2\left(y_{n-2}+\frac{3}{2}\right)^{2}\right]^{2}=2^{1+2}\left(y_{n-2}+\frac{3}{2}\right)^{2^{2}}$
$$
=\cdots=2^{1+2+\cdots+2^{n-2}}\left(y_{1}+\frac{3}{2}\right)^{2^{n-1}}=2^{2^{n-1}-1}\left(\frac{5}{2}\right)^{2^{n-1}}=\frac{1}{2} \cdot 5^{2^{n-1}} \text {, }
$$

and by mathematical induction, it is easy to prove that $x_{n}>0$, so
$$
x_{n}=\sqrt{y_{n}}=\sqrt{\frac{1}{2}\left(5^{2^{n-1}}-3\right)} .
$$