Constructing a local non-Noetherian domain with spectrum $\{(0), \mathfrak{m} \}$.

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); these are known as Hahn series and they form 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.