If $\mathrm{char} \, k \neq 2, 3$, then $R = k[x, y]/(y^2 - x^3 - 10)$ is a Dedekind domain

This problem is from the book Álgebra comutativa em quatro movimentos by Borges and Tengan.

Let $k$ be a field of characteristic not equal to $2$ or $3$. Let $f(x, y) = y^2 - x^3 - 10$ and consider the ring $R = k[x, y]/(f(x, y))$. Show that $R$ is a Dedekind domain.

Edit. We also have to assume $\mathrm{char} \, k \neq 5$.

This is what I've tried. We must show $R$ is (1) Noetherian, (2) one-dimensional and (3) normal. Condition (1) follows from the fact that $k[x, y]$ is Noetherian, plus correspondence theorem. For the other two conditions, I know a criterion when $k$ is algebraically closed.

Suppose $\bar k = k$ -- we write $k = \mathbb C$. It is easy to show that $\dim R = 1$ using the characterization of $\mathrm{Spec} \, \mathbb C[x, y]$. Namely, then any maximal chain of primes is of the form $(\bar 0) \subsetneq (\bar x - a, \bar y -b )$ with $f(a, b) = 0$. For normality, we have this useful corollary of Nakayama's lemma.

Let $f \in \mathbb C[x, y]$ be an irreducible polynomial and let $R = \mathbb C[x, y]/(f(x, y))$. Then, for $\mathfrak m = (\bar x - a, \bar y - b) \subset R$, $$\dim_{\mathbb C} \frac{\mathfrak{m} R_{\frak m}}{(\mathfrak{m}R_{\frak m})^2} = \begin{cases} 2, &\text{if } \partial_xf(a, b) = \partial_y f(a, b) = 0, \\ 1, &\text{otherwise.}\end{cases}$$ In particular, this says that $\mathfrak m R_{\mathfrak m}$ is principal iff $(a, b)$ is nonsingular.

But a one-dimensional noetherian ring is normal iff $\mathfrak m R_{\mathfrak m}$ is principal. So we are morally done. My question is how to adapt this to fields that are not algebraically closed. I thought of trying to explictly compute $\mathrm{Spec} \, R$ maybe using some isomorphism $R \cong k[t]$, but I couldn't develop that further. I appreciate any help.