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Let $\mathcal{P}$ be the set of primes not exceeding $10^{100}$. For each $p \in \mathcal{P}$, let $e_p = \max_x \nu_p(f(x))$ and let $c_p = \underset{x}{\operatorname{argmax}} \nu_p(f(x))$.
We show that this is good enough to compute all values of $x$, by looking at the exponent at each individual prime.

Claim — For any $p \in \mathcal{P}$, we have
$$\nu_p(f(x)) = \min(\nu_p(x - c_p), e_p).$$
Proof. Note that for any $x$, we have
$$\text{gcd}(f(c_p), f(x)) = \text{gcd}(f(c_p), x - c_p).$$
We then take $\nu_p$ of both sides and recall $\nu_p(f(x)) \le \nu_p(f(c_p)) = e_p$; this implies the result. $\square$

This essentially determines $f$, and so now we just follow through. Choose $n$ and $m$ such that
$$n = \prod_{p \in \mathcal{P}} p^{e_p} \\ m \equiv -c_p \pmod{p^{e_p}} \quad \forall p \in \mathcal{P}$$
the latter being possible by Chinese remainder theorem. Then, from the claim we have
$$\begin{aligned} f(x) &= \prod_{p \in \mathcal{P}} p^{\nu_p(f(x))} = \prod_{p \nmid n} p^{\min(\nu_p(x-c_p), e_p)} \\ &= \prod_{p \mid n} p^{\min(\nu_p(x+m), \nu_p(n))} = \text{gcd}(x+m, n) \end{aligned}$$
for every $x \in \mathbb{Z}$, as desired.