Spectrum of a ring is irreducible if and only if nilradical is prime (Atiyah-Macdonald, Exercise 1.19)

I think the open sets definition of irreducibility is easier to work with. You should show these useful facts about $Spec(A)$: the sets $D(f) = \{\mathfrak{p} \in Spec(A) : f \notin \mathfrak{p}\}$ form a basis of the topology of the spectrum, and $D(f) \cap D(g) = D(fg)$. Then we suppose that $f \notin Nil(A)$ and $g \notin Nil(A)$. This means that $D(f)$ and $D(g)$ are then nonempty open sets, and so if $Spec(A)$ is irreducible, their intersection $D(fg)$ is nonempty.