Answer: $f(x) = \epsilon x^d + c$, where $\epsilon \in \{-1, 1\}$, the positive integer $d$ is a divisor of $n$, and $c$ is an integer.
All functions $f$ satisfying the conditions of the problem are exactly the above.

Let the function $f$ be a solution satisfying the conditions of the problem. For any integer $n$, define the function $g(x) = f(x) + n$, which also satisfies the conditions of the problem. We may assume $f(0) = 0$.

For any prime $p$, taking $(x, y) = (p, 0)$ gives $f(p)|p^n$. Since there are infinitely many primes, there exist an integer $d$ and $\epsilon$ with $0 \le d \le n$, $\epsilon \in \{-1, 1\}$ such that there are infinitely many primes $p$ satisfying $f(p) = \epsilon p^d$. Denote the set $P$ as follows:
$$P = \{p \text{ is prime} \mid f(p) = \epsilon p^d\}.$$
Since the function $g$ satisfies the conditions of the problem if and only if $(-g)$ also satisfies the conditions of the problem, we may assume $\epsilon = 1$.

Exclude the case $d=0$. Assume $d \ge 1$ and $n = md + r$, where $m$ and $r$ are integers such that $m \ge 1$ and $0 \le r \le d-1$. Let $x$ be an arbitrary integer. For a prime $p$ in the set $P$, $(f(p) - f(x))|(p^n - x^n)$. Using the equality $f(p) = p^d$, we obtain
$$p^n - x^n = p^r(p^d)^m - x^n \equiv p^r(f(x))^m - x^n \equiv 0 \pmod{p^d - f(x)}.$$
Since $r < d$, for sufficiently large primes $p$ in $P$, we have
$$|p^r(f(x))^m - x^n| < p^d - f(x).$$
Therefore $p^r(f(x))^m - x^n = 0$. From this we can deduce: $r=0$ and $x^n = (x^d)^m = (f(x))^m$.
Since $m$ is odd, we get $f(x) = x^d$.