When is the prime spectrum of a ring the finite complement topology on a set? I have been studying the prime spectrum of different rings recently, and I have noticed that for many "nice" infinite rings, the prime spectrum is precisely the finite complement topology on some set along with a generic point(e.g., $\mathbb{Z}, \mathbb{C}[X]$). 
Now, if an infinite ring $A$ is a PID, I think I can outline a proof that the prime spectrum is a finite complement: for any finite subset $\{\langle p_i\rangle\}$ of the prime ideals, $X$, we can take $V(\{ \prod_i p_i\})$. Conversely, no infinite set of prime ideals can be closed, because that would require there to be an element of $A$ that is divisible by the infinitely many generators of these prime ideals, which is impossible, because $A$ is also a UFD. Adding the empty set and the whole space, we seem to have characterized the closed sets of Spec $A$, and so after we add a separated point to correspond to the zero ideal, we seem to be done. 
My question: is the above reasoning legitimate, and, if it is legitimate, are PID's the only time when a ring can have a prime spectrum a finite complement? If other rings can have such a prime spectrum, what sort of rings are they?

You are asking, I think, about rings $R$ whose spectrum $X=\operatorname {Spec}(R)$ has  the following property : 

$X$ has a special point $\eta$ and the closed subsets of $X$ are exactly $X$ and the finite subsets $\{x_1,...,x_n\}\subset X \;(n\in \mathbb N)$  of $X$ consisting of non-special points $x_i\neq \eta \:.$    

The noetherian rings $R$ with that property are exactly those satisfying :  

The  reduced ring $R_{\operatorname {red}=}R/\operatorname {Nil}(R)$ is a domain in which all non-zero primes are maximal. 

Examples
i) Dedekind domains $R$ qualify, for example PID's.
ii) Rings of the form $R=k[x,y]/\langle f(x,y)\rangle$, where $k$ is a field and $f(x,y)\in k[x,y]$ is an irreducible polynomial.
$X=\operatorname {Spec}(R)=V(f)\subset \mathbb A^2_k$ is then called a (maybe singular) affine plane irreducible algebraic curve.
iii) A slight modification  of the above example will produce non-reduced rings with the required properties: just take $R_n=k[x,y]/\langle (f(x,y))^n\rangle$ for some integer $n\geq 2$.
