Thinking Question 2 Given the sequence $\left\{x_{n}\right\}$ satisfies $x_{n+2}=\frac{x_{n+1}^{2}-4}{x_{n}}$, and $x_{1}=6, x_{2}=4$, find the general term $x_{n}$.
Given:
$$
\begin{array}{l}
x_{n+2} \cdot x_{n}=x_{n+1}^{2}-4, \\
x_{n+1} \cdot x_{n-1}=x_{n}^{2}-4,
\end{array}
$$
(1) - (2) yields $x_{n+2} x_{n}-x_{n+1} x_{n-1}=x_{n+1}^{2}-x_{n}^{2}$, which simplifies to
$x_{n}\left(x_{n+2}+x_{n}\right)=x_{n+1}\left(x_{n+1}+x_{n-1}\right)$, or
$$
\frac{x_{n+2}+x_{n}}{x_{n+1}}=\frac{x_{n+1}+x_{n-1}}{x_{n}}=\frac{x_{n}+x_{n-2}}{x_{n-1}}=\cdots=\frac{x_{3}+x_{1}}{x_{2}}=2 \text {. }
$$

Therefore, $x_{n+2}-2 x_{n+1}+x_{n}=0$, which simplifies to
$x_{n+2}-x_{n+1}=x_{n+1}-x_{n}=x_{n}-x_{n-1}=\cdots=x_{2}-x_{1}=-2$, so
$$
x_{n}=6-2(n-1)=8-2 n \text {. }
$$

Clearly, when $n=4$, $x_{n}=4$, so the recurrence relation only holds for the first 4 terms. When $n=5$, the denominator in the recurrence relation is zero, which is undefined, hence $x_{n}=8-2 n(1 \leqslant n \leqslant 4)$.