In any commutative ring with unity, prime ideals are proper (by definition), irreducible, and radical. As noted in comments and answers above, the converse is true: proper, irreducible, and radical ideals are prime. However, this can be done without a "sledgehammer"; it does not require Noetherianness, primary ideals, etc. More precisely:
Claim: Let $R$ be a commutative ring with unity. Let $I \subsetneq R$ be a proper ideal. Assume that $I$ is irreducible (if $I = J_1 \cap J_2$, then $I=J_1$ or $I=J_2$) and $I$ is radical (if $x^n \in I$ for any $n \geq 1$, then $x \in I$). Then $I$ is prime ($I \neq (1)$, and if $ab \in I$, then $a \in I$ or $b \in I$).
Remark: Note that the unit ideal $I=(1)=R$ is irreducible and radical, but not prime. In $R=\mathbb{Z}$, $I=(4)$ is proper and irreducible, but not prime; $I=(6)$ is proper and radical, but not prime. So, no two out of $\{$proper, irreducible, radical$\}$ are sufficient to get primeness—which is no surprise, since primeness implies proper, irreducible, and radical (all three).
Anyway, the hypotheses are necessary, let us proceed to the proof that they are sufficient.
Proof of Claim: By contrapositive. Let $I$ be a proper ideal, that is $I \neq (1)$. Assume $I$ is radical but not prime; our goal is to show that $I$ is not irreducible. In general there are two ways that $I$ can fail to be prime: either $I=(1)$, or there are some elements outside $I$ whose product is in $I$. We are assuming that $I$ is a proper ideal, $I \neq (1)$, so it must be the second possibility.
So there are some elements $a,b \in R$ such that $ab \in I$ while $a,b \notin I$. Consider the ideals $I+(a)$ and $I+(b)$. Certainly $I \subseteq (I+(a)) \cap (I+(b))$. For the reverse inclusion, suppose $x \in (I+(a)) \cap (I+(b))$. Write $x = i_1 + r_1 a = i_2 + r_2 b$ for some $i_1,i_2 \in I$ and $r_1,r_2 \in R$. Then $ x^2 = (i_1 + r_1 a)(i_2 + r_2 b) \in I $ since, upon expansion, three of the terms involve $i_1$ or $i_2$ (or both) and the fourth term has $ab$, which is in $I$. Since $I$ is radical, $x \in I$. This proves $I = (I+(a)) \cap (I+(b))$. Since $a \notin I$, $I \neq I+(a)$; similarly $I \neq I+(b)$. Therefore, $I$ is not irreducible. $\square$