How can I prove that polynomial ring $\mathbb{Z}[x]$ is not an integral domain ?

I was thinking that $\mathbb{Z}[x]$ is not a field so it is will not form integral domain as every finite integral domain is a field but here $\mathbb{Z}[x]$ contain infinitely many element so ,it will not form field .....

Is my thinking is correct or not please tell and verify me ?
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$\mathbb Z [x] $ can be considered a subring of $\mathbb Q [x] $. $\mathbb Q $ is a field, so $\mathbb Q [x] $ is an integral domain, and so $\mathbb Z [x] $ must be too (as any zero divisors in $\mathbb Z [x] $ will also be zero divisors in $\mathbb Q [x] $).
You don't explicitly say what $x$ is. If, as others seem to assume, $x$ is an indeterminate (an unknown variable), then of course $\mathbb{Z}[x]$ is an integral domain (no "zero divisors"). 

On the other hand one can construct a simple overring of the integers $\mathbb{Z}[x]$ in which there are zero-divisors. For example one might take $\mathbb{Z}[x] \simeq \mathbb{Z}[X]/\langle X^2 \rangle$, in which case $x^2 = 0$ is a nilpotent element of the overring.