Every open subset is affine

Question

I have the following question: given a ring $R$, what conditions should verify $R$, such that every open subset of $\text{Spec}(R)$ is affine?

Answer 1

Motivation: consider the ring $R=\mathbf{C}[x,y]$ of polynomial functions on $\mathbf{C}^2$. The union of the two principal open subschemes with coordinate rings $R[x^{-1}]$ and $R[y^{-1}]$ is not affine by Hartog's theorem: the ring of functions on it is $\mathbf{C}[x,y]$. Geometrically, at the level of closed points this open subscheme is the punctured plane $\mathbf{C}^2 \setminus \{(0,0) \}$.

Something similar happens at least whenever your ring $R$ is Noetherian, normal, and of Krull dimension at least $2$: fixing a maximal ideal $m$ of $R$ that is not generated by one element we have a non-affine open subset $$\bigcup_{f \in m} \mathrm{Spec}(R[f^{-1}])=\mathrm{Spec}(R) \setminus \{ m \}.$$ This subset is not affine because, by the algebraic version of Hartog's lemma (see e.g. Ravi Vakil's book), the space of sections of $\mathcal{O}_{\mathrm{Spec}(R)}$ over it is just $R$.