If $f \in \mathbb{Z}[x]$ is such that $p \mid f(p)$ for all primes $p$, then $x \mid f(x)$ in $\mathbb{Z}[x]$. This follows by writing $f(x) = \sum \limits_{i=0}^d c_i x^i$ and noting that $c_0 \equiv 0$ modulo $p$ for every prime $p$ implies that $c_0 = 0$.
There are perhaps several ways to generalize this to several variables; I am particularly interested in the following. (See the re-reformulation below.)
Let $f \in \mathbb{Z}[x_1,\ldots,x_n]$ be such that for every set of distinct primes $p_1,\ldots,p_n$, there exists an $i \in \{1,\ldots,n\}$ such that $f(p_1,\ldots,p_n) \equiv 0$ modulo $p_i$. Is it true that $x_1 \cdots x_n$ divides $f(x_1,\ldots,x_n)$ in $\mathbb{Z}[x_1,\ldots,x_n]$?
Edit: It was pointed out to me that $f(x_1,\ldots,x_n) = x_1$ answers the question above in the negative. I wish to add that $f \in \mathbb{Z}[x_1,\ldots,x_n]$ should be strictly contained in $\mathbb{Z}[x_1,\ldots,x_n]$ in the sense that all of the variables are present. Note that $f(x_1,\ldots,x_n) = x_1$ lies in $\mathbb{Z}[x_1]$ for which the answer is proved affirmative above.
Edit$^2$: Apparently $f(x_1,\ldots,x_n) = x_1 (x_1 + \cdots + x_n)$ answers the reformulated question in the negative. For my purposes, the following would suffice.
Let $f \in \mathbb{Z}[x_1,\ldots,x_n]$ be such that for every set of distinct primes $p_1,\ldots,p_n$, there exists an $i \in \{1,\ldots,n\}$ such that $f(p_1,\ldots,p_n) \equiv 0$ modulo $p_i$. Is it true that there exists an $i \in \{1, \ldots, n\}$ such that $x_i \mid f(x_1,\ldots,x_n)$ in $\mathbb{Z}[x_1,\ldots,x_n]$?