Let $R=\mathbb{Z}[\frac{-1+\sqrt{-19}}{2}]=\mathcal{O}_K$, where $K=\mathbb{Q}(\sqrt{-19})$.
A classical way to prove the fact that $R$ is a PID is to establish that for all $a,b\in R,b\neq 0$, there exists $q,r\in R$ such that $a=bq+r$ or $2a=bq+r$, where $\vert r\vert^2<\vert b\vert^2$.
I would like to find a proof of the fact that $R$ is a PID more in the spirit of "class group computations", using Minkowski's bound in disguise. Since I would like to present such a proof to my students (which have no knowledge of the classical results of algebraic number theory), I would like to avoid the whole machinery of algebraic number theory, so the question is:
Question. Given an ideal $I$ of $R$, is there an ad hoc and down-to-earth proof of the existence of $z\in K^\times$ such that $zI$ is an ideal of $R$ satisfying $\vert R/zI\vert\leq 2$.
Once we have this, it is easy to see that $zI=R$ or $2R$ (using maximality of $2R$) and we are done.
If it helps, I know an ad hoc proof of the fact that $I=\mathbb{Z}m+\mathbb{Z}z_0$, for some $m\in\mathbb{Z}$ and $z_0\in R$. Note that we may assume that $z_0\notin \mathbb{Z}$, and so $m, \mathbb{z}_0$ is a $\mathbb{Z}$-basis of $I$.