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A DVR necessarily has spectrum ${ 0, mathfrak{m} }$, but a DVR is also necessarily noetherian. Can we find an example of a non-noetherian (thus non DVR) ring, say $R$, with the spectrum ${ 0, mathfrak{m} }?$

Clearly we need $R$ a domain since the zero ideal is in the spec. We also have $mathfrak{m} neq 0$ since then $R$ would be a field, thus noetherian but even if it didn’t follow that $mathfrak{m} neq 0$ I would require this so it’s interesting. Also, krull dimension is obviously 1 since we have one chain of prime ideals of height 1.

I have tried the standard non-noetherian things like starting with a polynomial ring in countably many variables and quotienting and localizing, but so far no luck. I always accidentally land on a field instead of a 2 point spectrum. I also haven’t been able to come up with a way to see this from an algebro-geometric side of things.

My commutative algebra skills are moderate, I went about halfway through Atiyah Macdonald. I’m sure someone with stronger comm alg intuition will see an easy construction?

To expand on Bernard's answer, let $k$ be a field and consider the ring $R = k[[x^{mathbb{R}_{ge 0}}]]$ of formal power series of the form $sum_{r in S} c_r x^r$ where $r in mathbb{R}_{ge 0}$ and $S subseteq mathbb{R}_{ge 0}$ is well-ordered (this guarantees that multiplication is well-defined). This is a valuation ring.

For $f in R$ a nonzero power series, write $nu(f) in mathbb{R}_{ge 0}$ for the smallest exponent of a nonzero term in $f$. If $I$ is any nonzero ideal of $R$, then if some nonzero $f in I$ has valuation $nu(f)$, then multiplying by every element of $R$, and observing that nonzero every $g in R$ can be written as $x^{nu(g)}$ times a unit, we see that $I$ contains every element of $R$ with valuation $ge nu(f)$. This means $I$ is determined by the valuations of its nonzero elements, and the set of all such valuations is an upward-closed subset of $mathbb{R}_{ge 0}$ (upward-closed means if $x in S$ and $y ge x$ then $y in S$). There are two infinite families $[r, infty)$ and $(r, infty)$ of such subsets, so the ideals of $R$ come in two infinite families, namely

$$I_r = { f in R : nu(f) ge r }, r in mathbb{R}_{ge 0}$$

and

$$J_r = { f in R : nu(f) > r }, r in mathbb{R}_{ge 0}$$

(together with the zero ideal, which you can think of as $I_{infty}$). These ideals are totally ordered and in particular show that $R$ is very far from Noetherian. Note that $I_r = (x^r)$ is principal but $J_r$ is not even countably generated.

The corresponding quotients $R/I_r$ and $R/J_r$ contain nontrivial nilpotents, and hence $I_r$ and $J_r$ are not prime, except for $J_0$, which is the unique maximal ideal, and $I_0$, which is the unit ideal. So $R$ has exactly two prime ideals, $(0)$ and $J_0$, as desired.

A non-discrete valuation ring of height $1$ is such an example.