4. Given that $p$ is a prime number. Find all integer-coefficient polynomials $P(x)$ such that:
(1) $P(x)>x\left(x \in \mathbf{Z}_{+}\right)$;
(2) If the sequence $\left\{p_{n}\right\}(n \geqslant 0)$ satisfies:
$$
p_{0}=p, p_{n+1}=P\left(p_{n}\right)(n \in \mathbf{N}) \text {, }
$$

then for each $m \in \mathbf{Z}_{+}$, there exists $l \geqslant 0$, such that $m \mid p_{l}$.
4. $P(x)=x+1$ or $P(x)=x+p$.

First, prove that there exists $c \in \mathbf{Z}_{+}$, such that $P(x)=x+c$.
In fact, by condition (1),
$p_{n+1}>p_{n}(n \in \mathbf{N})$.
Thus, $\left\{p_{n}\right\}$ is a monotonically increasing and unbounded sequence.
For any positive integer $k$, by condition (2), there exists $l \geqslant 0$, such that $\left(p_{k}-p\right) \mid p_{l}$. We call a natural number $l$ with this property "k-good". Take the smallest k-good $l$, still denoted as $l$.
We will prove: $l \leqslant k-1$.
If not, $l \geqslant k$. We can prove that $l-k$ is also k-good.
Since $P$ is a polynomial, for any $x, y \in \mathbf{Z}_{+}$, and $x \neq y$, we have
$(x-y) \mid(P(x)-P(y))$.
For $i>j,\left(p_{i}-p_{j}\right) \mid\left(p_{i+1}-p_{j+1}\right)$, then $\left(p_{k}-p\right) \mid\left(p_{k+1}-p_{1}\right)$,
$\left(p_{k+1}-p_{1}\right) \mid\left(p_{k+2}-p_{2}\right)$,
......
$\left(p_{l-1}-p_{l-k-1}\right) \mid\left(p_{l}-p_{l-k}\right)$.
Thus, $\left(p_{k}-p\right) \mid\left(p_{l}-p_{l-k}\right)$.
Therefore, $\left(p_{k}-p\right) \mid p_{l-k}$, which contradicts the minimality of $l$.
Hence, $l \leqslant k-1, p_{k} \leqslant p+p_{k-1}$.
Therefore, for $x=p_{k-1}(k=1,2, \cdots)$, we have
$$
P(x) \leqslant x+p \text {, }
$$

i.e., for $x=p_{k-1}(k=1,2, \cdots)$, we have
$$
P(x)-x \leqslant p \text {. }
$$

Since $\left\{p_{n}\right\}$ is unbounded, $P(x)$ must be a monic first-degree polynomial (by condition (1), $P(x)$ is not a 0-degree polynomial, nor is it a polynomial with a negative leading coefficient), i.e., there exists $c \in \mathbf{Z}_{+}, P(x)=x+c$.
By the condition, there exists a 1-good $l$, take the smallest $l$, then, $l=0$.
Thus, $\left(p_{1}-p\right) \mid p$
$\Rightarrow p_{1}=p+1$ or $p_{1}=2 p$
$\Rightarrow P(p)=p+1$ or $P(p)=2 p$
$\Rightarrow c=1$ or $c=p$.
Therefore, $P(x)=x+1$ or $P(x)=x+p$.
It is easy to verify that these two polynomials meet the requirements.