Let $k$ be a non algebraically closed field with $i \not \in k$; equivalently the polynomial $T^2+1 \in k[T]$ is irreducible over $k$.
How to prove or disprove that for the ring $R:=k[y,z]/(1-y^2+z^2)$ every localization $R_{\mathfrak{p}}$ with repect every maximal prime ideal $\mathfrak{p}$ in $R$ (which means only that we require $\mathfrak{p} \neq \bar{(0)}$) is a UFD (unique factorization domain).
I would like also to highlight that this question arose naturally from this one: Stalks of Plane Conic $C \cong \mathbb{P}^1$ are UFD
In those case the field $k$ was assumed to be alg closed and we were able to identify the subvarieties $V_+(XZ-Y^2) $ and $V_+(y_0^2-y_1^2 +_2^2)$ of $\mathbb{P}^1_k$ by the transformation $X:= y_0+iy_2, Z:= y_0-iy_2$ and $Y:=y_1$. In that case we could do all our calculations on $V_+(XZ-Y^2) $, what turned out to be more easier.
Nevertheless, here for $k$ with $i \not \in k$ this transformation is invalid, and therefore we have to work with projective variety $V_+(y_0^2-y_1^2 +_2^2)$ or as in case above on it's affine chart $D_+(y_0) \cong \operatorname{Spec} \ k[y,z]/(1-y^2+z^2)$.
Expanding on user26857's comments....
We only require the assumption that the characteristic of $k$ is not $2$, which is a weaker assumption than $i \notin k$. This assumption is necessary: if $\operatorname{char}(k) = 2$ then $1-y^2 + z^2 = (1+y)^2 + z^2 = (1+y + z)^2$, hence $k[y,z]/(1 - y^2 +z^2)$ is not even reduced, and in particular cannot be locally a domain, let alone locally a UFD.
On the other hand if $0 \not= 2 \in k$, note that $k[y,z] = k[y-z, y+z]$, and $k[y,z]/(1 - y^2 + z^2) = k[y-z, y+z]/(1 - (y-z)(y+z))$. Thus taking $u = y-z, v = y+z$ your ring is just $k[u,v]/(1 - uv) \cong k[u,u^{-1}, v] \cong k[u,v]_u$, which is the localization of a UFD, and hence a UFD.