Three, (50 points) Let $f(x)$ be a polynomial with integer coefficients. For any prime $p$ and integers $u, v$, if $p \mid (uv + u + v)$, then $p \mid (f(u) f(v) - 1)$. In this case, $f(x)$ is called "good". Find all good $f(x)$.

保留源文本的换行和格式，直接输出翻译结果。
Three, first prove a lemma.
Lemma For any prime $p$ and integers $s, t$, if $p \mid (st-1)$, then $p \mid (f(s)f(t)-1)$, the integer-coefficient polynomial $f(x) = \pm x^n (n \in \mathbf{N})$.
Proof When $f(x)$ is a constant, $f(x) = \pm 1$;
When $f(x)$ is not a constant, let $\operatorname{deg} f(x) = n$, and set $g(x) = x^n f\left(\frac{1}{x}\right)$.
Then $g(x)$ is also an integer-coefficient polynomial. For a prime $p$, let $u^{-1}$ be the inverse of $u$ modulo $p$. Thus, by the lemma condition, for any integer $u$ satisfying $(u, p) = 1$, we have
$$
\begin{array}{l}
p \mid (f(u) f(u^{-1}) - 1) \\
\Rightarrow p \mid (u^n f(u) f(u^{-1}) - u^n). \\
\text{By } u^n f\left(\frac{1}{u}\right) \equiv u^n f(u^{-1}) (\bmod p) \\
\Rightarrow p \mid (f(u) g(u) - u^n).
\end{array}
$$

For a fixed integer $u$ and any prime $p$,
$$
p > |f(u) g(u) - u^n|,
$$

From equation (1), we get $f(u) g(u) - u^n = 0$.
Thus, for all integers $x$, $f(x) g(x) = x^n$.
Since $\operatorname{deg} f(x) = n$, we know $g(x) = \pm 1$.
Therefore, $f(x) = \pm x^n$.
The lemma is proved.
In the lemma, let $s = u + 1, t = v + 1$, we know the conclusion of the original problem is $f(x) = \pm (x + 1)^n (n \in \mathbf{N})$.