Find all real-coefficient polynomials $P(x)$ such that there exists an infinite sequence of integers $\left\{a_{n}\right\}$, all distinct, and satisfying
$$
P\left(a_{1}\right)=0, P\left(a_{k+1}\right)=a_{k}(k=1,2, \cdots) .
$$
【Analysis】From the previous proof process, $P(x)$ being a constant function still does not meet the requirements.

Below, we prove that when $\operatorname{deg} P \geqslant 2$, there are no polynomials that meet the requirements.
When $\operatorname{deg} P \geqslant 2$, $\lim _{x \rightarrow+\infty}\left|\frac{P(x)}{x}\right|=+\infty$.
Therefore, there exists $x_{0} \in \mathbf{Z}_{+}$, such that when $|x| \geqslant x_{0}$,
$|P(x)|>|x|$.
Since there are only finitely many integers with absolute value less than $x_{0}$, there exists $n_{0} \in \mathbf{Z}_{+}$, such that when $n \geqslant n_{0}$,
$\left|a_{n}\right| \geqslant x_{0}$.
Thus, when $n \geqslant n_{0}$,
$\left|a_{n}\right|=\left|P\left(a_{n+1}\right)\right|$
$>\left|a_{n+1}\right|=\left|P\left(a_{n+2}\right)\right|$
$>\left|a_{n+2}\right|=\cdots$
Therefore, the sequence of positive integers $\left\{\left|a_{n}\right|\right\}$ is strictly decreasing from the $n=n_{0}$ term onwards and has a lower bound $x_{0}$, which is impossible.

Thus, when $\operatorname{deg} P \geqslant 2$, there are no polynomials that meet the requirements, i.e.,
$\operatorname{deg} P=1$.
Let $P(x)=A x+B(A, B \in \mathbf{R}, A \neq 0)$. Then
$P\left(a_{1}\right)=0 \Rightarrow a_{1}=-\frac{B}{A} \in \mathbf{Z}$,
$P\left(a_{k+2}\right)=a_{k+1}, P\left(a_{k+1}\right)=a_{k}$
$\Rightarrow A=\frac{a_{k+1}-a_{k}}{a_{k+2}-a_{k+1}} \in \mathbf{Q}$.
Let $A=\frac{s}{t}(s, t \in \mathbf{Z}, s>0, t \neq 0, (s, t)=1)$.
Then $a_{k+2}-a_{k+1}$
$=\frac{t\left(a_{k+1}-a_{k}\right)}{s}=\cdots=\frac{t^{k}\left(a_{2}-a_{1}\right)}{s^{k}}$.
Since $(s, t)=1$, for any $k \in \mathbf{Z}_{+}$,
$s^{k} \mid\left(a_{2}-a_{1}\right) \Rightarrow s=1$
$\Rightarrow A=\frac{1}{t}(t \neq 0, t \in \mathbf{Z})$.
When $t=-1$,
$-a_{k+1}+B=a_{k},-a_{k+2}+B=a_{k+1}$
$\Rightarrow a_{k+2}=a_{k}$,

which is a contradiction.
When $t \neq-1,0$, $a_{k+1}=t a_{k}-B t$.
Let $D=B t$. Then $D=\frac{B}{A} \in \mathbf{Z}$.
Thus, $P(x)=\frac{x+D}{t}(t \in \mathbf{Z} \backslash\{0,-1\}, D \in \mathbf{Z})$.
At this point, $a_{1}=-D$.
If $D=0$, then it is easy to see
$a_{2}=-(t+1) D=0=a_{1}$,

which is a contradiction.
Therefore, $D \neq 0$, and for any $t \in \mathbf{Z} \backslash\{0,-1\}$, $D \in \mathbf{Z} \backslash\{0\}$, we get
$$
a_{k}=\left\{\begin{array}{ll}
\frac{-D\left(t^{k}-1\right)}{t-1}, & t \neq 1,0,-1 ; \\
-k D, & t=1,
\end{array}\right.
$$

where $k=1,2, \cdots$.
It is easy to see that $a_{k} \in \mathbf{Z}$, and the sequence $\left\{a_{k}\right\}$ has distinct terms. In summary, all polynomials that satisfy the conditions are
$$
P(x)=\frac{x+D}{t}(t \in \mathbf{Z} \backslash\{0,-1\}, D \in \mathbf{Z} \backslash\{0\}) \text {. }
$$

Generalization 2 Find all complex-coefficient polynomials $P(x)$ such that there exists an infinite sequence of integers $\left\{a_{n}\right\}$, where any two terms are distinct, and satisfy
$P\left(a_{1}\right)=0, P\left(a_{k+1}\right)=a_{k}(k=1,2, \cdots)$.
【Analysis】Let $P(x)=\sum_{j=0}^{n}\left(\alpha_{j}+\beta_{j} \mathrm{i}\right) x^{j}$, where $\alpha_{j}, \beta_{j} \in \mathbf{R}(j=0,1, \cdots, n), \alpha_{n} \neq 0$.

Then, by the problem statement, for any $k \in \mathbf{Z}_{+}, P\left(a_{k+1}\right)=$ $a_{k} \in \mathbf{R}$, i.e.,
$\sum_{j=0}^{n} \beta_{j} a_{k+1}^{j}=0$.
Thus, the polynomial $\sum_{j=0}^{n} \beta_{j} x^{j}=0$ has infinitely many distinct roots $a_{k+1}, a_{k+2}, a_{k+3}, \cdots$
Therefore, the polynomial $\sum_{j=0}^{n} \beta_{j} x^{j}$ is the zero polynomial.
Thus, for any $j \in\{0,1, \cdots, n\}, \beta_{j}=0$, i.e., $P(x)$ is a real-coefficient polynomial, reducing to the case in Generalization 1.