Let $p_1, \ldots, p_n$ be a finite sequence of nonconstant polynomials with integer coefficients. Does there exist a finite sequence of integers $x_1, \ldots, x_n$ such that the integers $p_1(x_1), \ldots, p_n(x_n)$ have a nontrivial common divisor?
Here's an argument that might be a bit too highbrow (it uses Chebotarev's density theorem).
Reduction 1. Your question is equivalent to: for $f_1, \ldots, f_n \in \mathbb Z[x]$, does there exist a prime $p$ such that all $f_i$ have a root in $\mathbb F_p$. Indeed, taking lifts $x_i \in \mathbb Z$ of roots $\bar x_i \in \mathbb F_p$ of $\bar f_i \in \mathbb F_p[x]$ gives $$f_i(x_i) \equiv 0 \pmod p$$ for all $i$, so the $f_i(x_i)$ have $p$ as common divisor.
Reduction 2. We may assume that all $f_i$ are irreducible. Indeed, a root (mod $p$) of a factor of $f_i$ gives a root of $f_i$ itself.
Proof of the assertion. Let $K_i = \mathbb Q[x]/(f_i)$ be the field extension defined by adjoining a root of $f_i$. Let $K$ be the Galois closure of the compositum of the $K_i$ (inside $\bar {\mathbb Q}$). By the Chebotarev density theorem, there exists a prime $p$ such that $(p)$ is completely split in $K$. Thus, all the $f_i$ have a root modulo $p$. $\square$
Remark. This actually proves more: there are infinitely many primes $p$ with this property (and their density is $1/[K:\mathbb Q]$). Moreover, there are many lifts $x_i$ of a root of $\bar{f}_i$, so we actually get many examples of such $x_i$. One can ask how many there are (I will not attempt this right now).
Here is a bit more pedestrian argument. We may assume that $p_i$ are irreducible. Take $\alpha_i$ a root of $p_i$; then $\mathbb Q[\alpha_1,\dots,\alpha_n]=\mathbb Q[\xi]$ by the primitive element theorem. Taking $q$ to be the monic minimal polynomial of $\xi$, we have $\alpha_i=r_i(\xi)$, so $q(x)\mid p_i(r_i(x))$. Now we may take any prime divisor $p$ of the values of $q$ which does not appear in the denominators of the coefficients of $q$ and $r_i$; $p$ divides appropriate values of $p_i$ as well.

$\begingroup$ It would help to say: we can choose $\xi$ so that the $r_i$'s all have integral coefficients. Then when $p$ is an appropriate divisor of $q(x_0)$, it also divides $p_i(r_i(x_0))$, and those are appropriate integer values of $p_i$. Without a careful choice of $\xi$, the $r_i(x_0)$ and $p_i(r_i(x_0))$ could be fractional. $\endgroup$– Matt F.Sep 10 '16 at 1:07

$\begingroup$ @Matt: That's why I speak on the prime divisors $p$ which do not divide the denominators of the coefficients (there exist such, because the values of $q$ have infinitely many prime divisors). I am not sure tha you can choose $\xi$ so that $q$ and all the $r_i$ have integer coefficients. $\endgroup$ Sep 10 '16 at 12:24

$\begingroup$ @IlyaBogdanov I don't quite see through the last step. You can assume $r_i\in\mathbb Z[x]$. Then $q$ would generally be in $\mathbb Q[x]$ rather than $\mathbb Z[x]$. If you write $p_i(r_i(x))=q(x)s_i(x)$, then generally $s_i\in\mathbb Q[x]$. Then you have to choose prime divisor of the numerator of $q(x)$ that is also coprime to the denominators of $s_i(x)$. Without any information on $s_i$, I can't see a way to show this without showing that $q(x)$ has roots in infinitely many $\mathbb F_p$, which is more or less Chebotarev's density theorem. $\endgroup$ Jan 13 '17 at 1:26

1$\begingroup$ @FanZheng: The fact that the values of a nonconstant polynomial with integer coefficients at integer points have infinitely many prime divisors is elementary and easy. Arguing indirectly, assume that $q_0$ is the constant term, and $p_1,\dots,p_n$ are all prime divisors of the values; surely $q_0\neq 0$. Then it suffices to plug in $x=(Nq_0p_1\dots p_n)^2$, where $N$ is large enough to ensure that the value is greater than $q_0$ (in absolute value). $\endgroup$ Jan 20 '17 at 23:59