Question 1: Does such a ring exist?
Note: The definition of a noetherian topological space is similar to that in rings or sets. Every descending chain of closed subsets stops after a finite number of steps
(Question 2: is this equivalent to saying that every descending chain of opens stops?)

It is easy to come up with examples when you keep in mind that $X$ and $X_{\text{red}}$ are homeomorphic: The topology is not changed when you divide out (all) nilpotent sections.

For example, $A = k[x_1,x_2,...] / (x_1^2,x_2^2,...)$ satisfies $A/\mathrm{rad}(A)=k$, so that $\mathrm{Spec}(A)$ consists of exactly one point, namely the maximal ideal $(x_1,x_2,...)$, which is not finitely generated.

In general, $\mathrm{Spec}(A)$ is a noetherian topological space iff $A$ satisfies the ascending chain condition for radical ideals.
