Notice that the elements of this ring are of the form $a + bx$.
To show that $x$ is prime, suppose $x$ divides a non-zero number $m$ :
- then $m = bx$ for some $b$.
- if $m = (c + dx)(c'+d'x)$, then without loss of generality (wlog) $c = 0$ and $d \neq 0$. Hence $x$ divides $c + dx = dx$. That is, $x$ is prime.
But $x$ is not ``irreducible'' since $x = x^2$ and $x$ is not a unit (to show it is not a unit, show that $x(a+bx) = 1$ cannot be solved.
However, using the reference in the comments above, this example would not apply depending on the used definition of irreducible. The correct definition is:
In a ring $R$ (with out without zero divisors) an element $a$ is irreducible if whenever $a = bc$ then either $(a) = (b)$ or $(a) = (c)$. (See abstract of ref given in the comments above ).
Using the above definition, then every prime element is irreducible:
Suppose $p$ is a prime element. If $p = p_1 p_2$ then wlog $p$ divides $p_1$ since $p$ is prime. We have that $(p_1) = (p)$ since $p_1 \in (p)$ ($p$ divides $p_1$) and $p \in (p_1)$ ($p_1$ divides $p$).
 Anderson and Chun, Irreducible Elements in Commutative rings with non-zero divisors, http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.442.1455&rep=rep1&type=pdf