Here are some thoughts on why such a proof might be hard to find (and interesting). They are not totally rigorous, but I think they give a different perspective, so might be worth something.
A natural way to provide an irreducible element whose ideal is not radical is to create non-trivial torsions in the class group (I will assume the ring is normal). Indeed, just from definition, you need such an element $x$, and some elements $y,z$ such that
$$xy=z^n$$ but $z\notin (x)$. Now, if $P=(x,z)$ happens to be prime, then by computing the Weil divisor corresponding to the Cartier divisor $(x)$, one gets $n[P]=0$ in the class group. But $[P]$ is not principal since $z\notin (x)$ by assumption and $x\notin (z)$ by irreducibility of $x$.
This is precisely what happened in the examples by Qiaochu ($\mathbb Z[-\sqrt{5}]$) and Gerry ($k[x,y,z]/(xy-z^2)$) (both have class group $\mathbb Z/(2)$).
So it seems to me the proof you want would rule out certain torsions in the class group without showing that the group is trivial (which means our ring is a UFD). Unfortunately, understanding torsions in $\text{Cl}(R)$, especially over arbitrary fields, is a hard problem (think about elliptic curves!)
Does that make sense?