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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$\Bbb Z[X]$ is actually an integral domain:

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Let $f,g \in \Bbb Z[X]$ with $fg=0$.

Convince yourself (prove via induction) that if the leading coefficient of $f$ is $m$ and that of $g$ is $n$, then that of $fg$ is $mn$.

However, since the leading coefficients of both sides must be equal, we must have $mn = 0$.

However, $m$ and $n$ are integers, so $mn=0$ implies $m=0$ or $n=0$.

The leading coefficient of a polynomial, by definition, is $0$ only when the polynomial itself is $0$, so we conclude $f=0$ or $g=0$.

To conclude, for any $f,g \in \Bbb Z[X]$, $fg=0$ implies $f=0$ or $g=0$, so $\Bbb Z[X]$ is an integral domain.
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.