Well, the title of my question says all, but let me give some context. Right now I'm writing some lecture notes on ring theory with a little of commutative algebra. I wrote a few results about integral extensions and that led me to define the algebraic integers $\overline{\Bbb Z}$ as the integral closure of the extension $\Bbb Z\subseteq \Bbb C$. Since I included a topic on factorization in integral domains where I defined what is a Bézout domain, it seems very natural to me to show that $\overline{\Bbb Z}$ is an example of a Bézout domain.
However, and here comes my problem, I haven't found an elementary proof of the above fact. The only reference that I have is a theorem in Kaplansky's book "Commutative rings". I checked the proof given in that book, but I can't understand it very well, moreover Kaplansky uses terminology from Algebraic Number Theory as "class group" and certainly I can't use any of that in my notes because they are aimed to cover a basic/intermediate course of ring theory.
In summary, is there a ring-theoretic way to prove that the ring of algebraic integers is a Bézout domain?
P.S. I'm aware of this post and the only answer uses Algebraic Number Theory, which as I said, I can't use.