(Completing my comments above to an answer. Probably one can simplify this quite a bit.)
EDIT. The previous version mistakenly identified the ideal $xA\cap R$ with $xR$. In fact, the maximal ideal $xA\cap R$ of $R$ is not finitely generated: it is generated by $\{xf\,:f\in k(y)\}$ and a finite subset does not suffice. Similarly, $xR$ is not a prime ideal: we have $xy^{-1}, xy\notin xR$ but their product $x^2\in xR$.
The ring has exactly two prime ideals, $(0)\subseteq \mathfrak{m}$ where $\mathfrak{m} = A\cap xk[x,y]$.
Let $A = k[x, y]_{(x)}$ is the local ring at the generic point of the $y$-axis. This is a discrete valuation ring with maximal ideal $\mathfrak{m}=(x)$ and residue field $A/\mathfrak{m} = k(y)$.
The ring $R$ in question is the preimage of $k\subseteq k(y)= A/\mathfrak{m}$ in $A$. In other words, it is the fiber product $R = A\times_{A/\mathfrak{m}} k$.
EDIT (following Anton's comment below): a better reference for the following two paragraphs is stacks.math.columbia.edu/tag/0D2G Lemma 0B7J: the underlying space of the spectrum of the fiber product of the form $A\times_{A/I} B$ is the pushout of the corresponding underlying topological spaces of spectra.
By Stacks Project, Tag 07RS https://stacks.math.columbia.edu/tag/07RS, $\operatorname{Spec} R$ is the pushout of $\operatorname{Spec}k \leftarrow \operatorname{Spec} A/\mathfrak{m} \to \operatorname{Spec} A$.
By Theorem 3.4 (and its proof) in Schwede's paper http://www-personal.umich.edu/~kschwede/SchemeWithoutPoints.pdf , we get that the underlying space of $\operatorname{Spec} R$ is the corresponding pushout in spaces. But $\operatorname{Spec} A/\mathfrak{m}\to \operatorname{Spec} k$ is a homeomorphism, and hence so is $\operatorname{Spec} A\to \operatorname{Spec} R$.