Conside the ring $<\mathbb{Z}_5[x],+,.>$.
Is the ideal $ \langle1 + 3x + 3x^2 + x^3\rangle $ a prime ideal? If we have a ring of integers then we can simply check that if product of $a$ and $b$ belongs to the ideal $P$ then either $a$ or $b$ should belong to $P$.
How can I check this for polynomials?
Note that $(1+x)^3 = 1+3x+3x^2+x^3 \in (1+3x+3x^2+x^3)$, but $1+x \notin (1+3x+3x^2+x^3)$ because all non-zero multiples of $1+3x+3x^2+x^3$ must be of degree at least $3$. Thus, $(1+3x+3x^2+x^3)$ cannot be prime.
In general, if $F$ is a field then $F[x]$ is a principal ideal domain, and in particular every non-zero prime ideal is maximal. Therefore, one can merely check for reducibility of $p(x)$ in $F$ to see if $(p(x))$ is prime or not. If $\deg p \leq 3$, then in fact it is enough to check if $p$ has a root in $F$ or not i.e. if $\exists x \in F,p(x) =0$.
In our case, with $p(x) = x^3+3x^2+3x+1$,we may consider the element $4 \in \mathbb Z_5$, and see that $p(4) = 0$. So $p$ is reducible mod $5$, and hence we have that it cannot be prime.
If $F$ is not a field, then $F[x]$ need not be a principal ideal domain, as shown by the example $(2,x)$ being non-principal over $\mathbb Z[x]$. Here, prime ideals are more difficult to characterize, because one cannot say what every ideal looks like in such a field(it may not be principal, but also may not have other nice properties), so it is more difficult to deduce conclusions from something being an ideal in this case.