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.
A material implication $P\to Q$ is a statement like "If $P$, then $Q$". For example, "If a ring $R$ is finite and integral domain, then it is a field". 

To every material implication $P\to Q$, we can consider the converse $Q\to P$, the inverse $\neg P\to\neg Q$, and the contrapositive $\neg Q\to\neg P.$

If an implication is true, then so is its contrapositive; they are tautologically equivalent. However the converse and inverse are not equivalent and may fail to be true. Consider that "All men are mortal" is true, but "all mortals are men" is not. Assuming the converse of a true implication is also true is a variation of a logical fallacy sometimes known as affirming the consequent. Assuming the truth of the inverse implication is known as "denying the antecedent".

So if $P$ is the statement "$R$ is a finite integral domain", and $Q$ is the statement "$R$ is a field", the converse $Q\to P$ is the statement "All fields are finite integral domains", the inverse $\neg P\to\neg Q$ is the statement "if a ring is not a finite integral domain, it is not a field", and the contrapositive $\neg Q\to \neg P$ is the statement "if a ring is not a field, then it is not a finite integral domain." 

So in our case, if it is true that all finite integral domains are fields, it is also true that any ring which is not a field is not a finite integral domain.

What is *not* true is that any ring not a finite integral domain is therefore a field. It is also not the case that any ring that is not finite must not be an integral domain.

To the point, $\mathbb{Z}[x]$ is certainly an integral domain. See if you can prove that for $R$ an integral domain, $R[x]$ is also an integral domain.