Wolmer Vasconcelos  gave the following classification theorem about commutative local rings of global dimension two:
In this post we shall prove a small part of this theorem: that if $A$ is a commutative local ring of global dimension two and $M$ is a principal ideal, then $A$ is a valuation domain (i.e. for all $a,b\in A$ either $a | b$ or $b | a). As always, we use the term local ring to mean a commutative ring with a unique maximal ideal.
Before we prove this theorem, let us warm up by discussing the simpler cases of global dimension zero and one.
Rings of (left) global dimension one are called (left) hereditary. (Left) hereditary is the same as saying that every (left) ideal is projective. If a commutative local ring has global dimension one, then it must be a domain. Indeed, annihilator ideals are projective, and hence free. And, it is well-known that the commutative domains of global dimension one are exactly the Dedekind domains. Therefore, if $A$ is a commutative local ring of global dimension one, then it is not only a valuation domain but a discrete valuation domain.
We claim that $A$ is a GCD-domain. For suppose $a,b\in A$ are not both zero. We will calculate the GCD of $a$ and $b$. Consider the map $A^2\to (a,b)$ given by $(x,y)\mapsto ax – by$. Let $K$ be the kernel of this map so that we have an exact sequence
$$0 \to K\to A^2\to (a,b).$$
Then $K$ is a free module of rank one generated by an element $(\alpha,\beta)\in A^2$. Since $(b,-a)\in K$ we can write $(b,-a) = \delta (\alpha,\beta)$. Now it is easy to see that $\delta$ is in fact the GCD of $a$ and $b$.
Let $a,b\in A$ and let $\delta$ be their GCD. Write $a = d\alpha$ and $b = d\beta$ for some $\alpha,\beta\in A$. If either $\alpha$ or $\beta$ is a unit, then we are done, since this implies that either $a | b$ or $b | a$. If this were not true, then it would be the case that $d | \alpha$ and $d | \beta$, showing that $d\delta | a$ and $d\delta | b$. But $d\delta$ does not divide $\delta$ because $A$ is a domain, and this contradicts the fact that $\delta$ is the GCD of $a$ and $b$. QED
As we mentioned in the beginning, the conclusion also holds if $M$ is not finitely generated, but requires a little more work that would have made this post too long. But, if $M$ is finitely generated, Vansconcelos' theorem says that $M$ must be generated by a regular sequence of two elements $a,b\in M$. This means that $A$ cannot be a valuation domain! Indeed, $a | b$ would imply that $a$ would be a zero divisor on $R/b$.
For Noetherian local rings, this result becomes much easier because such rings of finite global dimension are automatically regular, and thus their maximal ideals must be generated by a regular sequence of two elements. However, this shows that Vasconcelos' theorem is also cool in that even if $A$ is not Noetherian, if $M$ is finitely generated and not principal, then it has a two-element generating set.
 Vasconcelos, Wolmer V. The local rings of global dimension two. Proc. Amer. Math. Soc. 35 (1972), 381-386.