Example 9 Let the sequence $\left\{x_{n}\right\}$ satisfy: $x_{1}=2, x_{2}=3$, and
$$
\left\{\begin{array}{l}
x_{2 m+1}=x_{2 m}+x_{2 m-1}, m \geqslant 1, \\
x_{2 m}=x_{2 m-1}+2 x_{2 m-2}, m \geqslant 2 .
\end{array}\right.
$$

Find the general term formula of the sequence $\left\{x_{n}\right\}$.
Solve: From the recurrence relation, when $m \geqslant 2$, we have $x_{2 m+1}=x_{2 m}+x_{2 m-1}=2 x_{2 m-1}+2 x_{2 m-2}$,
and $x_{2 m-2}=x_{2 m-1}-x_{2 m-3}$,
so $x_{2 m+1}=4 x_{2 m-1}-2 x_{2 m-3}$.
Let $a_{m}=x_{2 m-1}$, then $a_{m+1}=4 a_{m}-2 a_{m-1}, m \geqslant 2$.
Similarly, let $b_{m}=x_{2 m}$, we can also get $b_{m+1}=4 b_{m}-2 b_{m-1}, m \geqslant 2$.
The sequences $\left\{a_{m}\right\}$ and $\left\{b_{m}\right\}$ have the same recurrence relation, whose characteristic equation is $\lambda^{2}-4 \lambda+2=0$,
solving which gives the characteristic roots $\lambda=2 \pm \sqrt{2}$. Thus,
$$
\begin{array}{l}
a_{m}=\alpha_{1}(2+\sqrt{2})^{m}+\beta_{1}(2-\sqrt{2})^{m}, m=1,2, \cdots \\
b_{m}=\alpha_{2}(2+\sqrt{2})^{m}+\beta_{2}(2-\sqrt{2})^{m}, m=1,2, \cdots
\end{array}
$$

where $\alpha_{1}, \beta_{1}, \alpha_{2}, \beta_{2}$ are undetermined constants. Furthermore, by
$$
\begin{array}{l}
a_{1}=x_{1}=2, a_{2}=x_{3}=x_{2}+x_{1}=5, \\
b_{1}=x_{2}=3, b_{2}=x_{4}=x_{3}+2 x_{2}=11,
\end{array}
$$

we can calculate $\quad \alpha_{1}=\frac{1}{4}(3-\sqrt{2}), \beta_{1}=\frac{1}{4}(3+\sqrt{2})$,
$$
\alpha_{2}=\frac{1}{4}(1+2 \sqrt{2}), \beta_{2}=\frac{1}{4}(1-2 \sqrt{2}) .
$$

Therefore, when $n=2 m-1$, $x_{n}=\frac{1}{4}(3-\sqrt{2})(2+\sqrt{2})^{m}+\frac{1}{4}(3+\sqrt{2})(2-\sqrt{2})^{m}$; when $n=2 m$, $x_{n}=\frac{1}{4}(1+2 \sqrt{2})(2+\sqrt{2})^{m}+\frac{1}{4}(1-2 \sqrt{2})(2-\sqrt{2})^{m}$.