If the non-noetherianness of the ring is hidden inside the nilradical, then $\mathrm{Spec}$ won't see it.

Let $k$ be any ring, and let $V$ be a free $k$-module of infinite rank. Consider $R=k\oplus V$, and turn it into a ring by defining $$(a,v)\cdot(b,w)=(ab,aw+bv).$$ (Representation-people call this a *trivial extension*) Then $R$ is not-noetherian, because every $k$-submodule of $V$ is an ideal in $R$. Yet $V$ is contained in the nilradical of $R$: if you look at $\mathrm{Spec}\;R$ and at $\mathrm{Spec}\;k$, you'll see that they are very similar.

As for your second question: no. If $k$ is an infinite field, then $\mathrm{Spec}\;k[X]$ is noetherian, yet you'll easily find a decreasing chain of open sets init which does not stop. (What examples did you consider before asking the question? :) )