Example 14 (2005 National Team Training Question) Find all integer-coefficient polynomials $f(x)$ such that for all positive integers $n$, we have $f(n) \mid 2^{n}-1$.

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Assume $f(x)$ is not a constant, without loss of generality, let the leading coefficient of $f(x)$ be positive, then there exists a positive integer $N$, such that when $x \geqslant N$, $f(x) \geqslant 2$.
Take any positive integer $n, n \geqslant N$, take a prime factor $p$ of $f(n)$, since $f(n) \mid 2^{n}-1$, so $2^{n} \equiv 1(\bmod$ $p$.
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Also, since $f(n+p) \equiv f(n) \equiv 0(\bmod p), f(n+p) \mid 2^{n+p}-1$, so $2^{n+p} \equiv 1(\bmod p)$, thus $2^{p} \equiv 1(\bmod p)$.
By Fermat's Little Theorem, $2^{p} \equiv 2(\bmod p)$, thus $1 \equiv 2(\bmod p)$, a contradiction!
Therefore, $f(x)$ is a constant, let $f(x)=a$.
Since $f(1) \mid 2^{1}-1$, i.e., $a \mid 1$, we know $a= \pm 1$. Therefore, the integer coefficient polynomials sought are
$f(x)=1$ and $f(x)=-1$.