Let $X$ be a proper ring variety over a field $k$. Note first that $X$ is geometrically connected since it has a $k$-rational point ($0$ and $1$ are both rational points). Let $Y$ be the base-change of $X$ to an algebraic closure $\overline{k}$. Note that $Y_{red}$ is a subring of $Y$ (the ring operations $Y\times Y\to Y$ restrict to morphisms $Y_{red}\times Y_{red}\to Y_{red}$ since $Y_{red}\times Y_{red}$ is reduced). Since $Y$ is connected, so is $Y_{red}$, and it follows that $Y_{red}$ is irreducible (since its additive group structure makes it homogeneous). If we show that $0=1$ in $Y_{red}$, that will imply $0=1$ in $Y$, and so $0=1$ in $X$ as well so $X$ is the trivial ring variety.
The upshot of all of this is that $Y_{red}$ is a proper ring variety over $\overline{k}$ and it suffices to show that $Y_{red}$ is trivial. In other words, we may replace $X$ with $Y_{red}$ and assume that $k$ is algebraically closed.
Now consider the morphism $f:X\times X\to X\times X$ given by $f(x,y)=(xy,y)$. Since we are assuming $k$ is algebraically closed, $X\times X$ is irreducible. Note that the fiber of $f$ over $(1,1)$ is just $\{(1,1)\}$. It follows that the generic nonempty fiber of $f$ is $0$-dimensional and so the image of $f$ has the same dimension as $X\times X$. Since $X\times X$ is proper the image of $f$ is closed and since $X\times X$ is irreducible the image is dense, and so $f$ is surjective. In particular, there exist $x,y\in X$ such that $f(x,y)=(1,0)$. This implies $y=0$ and so $1=xy=0$. Thus $0=1$ in $X$ and $X$ is the trivial ring variety.
(This argument does not use the full ring structure of $X$, but merely uses the fact that $X$ has a multiplication operation with an element $1$ such that $x1=x$ for all $x$ and an element $0$ such that $x0=0$ for all $x$.)
First, we can replace $\overline{k}$ with a finite extension $K$ of $k$.
We know that $\mathcal{F}=\mathrm{Hom}_k(-,Y)$ is a fpqc-sheaf. We know that there is some $\phi’ \in \mathcal{F}(X_K)$ (which is the composition of $\overline{\phi}:X_K \rightarrow Y_K$ and $Y_K \rightarrow Y$).
As $X_K \rightarrow X$ is a fpqc cover, $\phi’$ comes from a morphism $\phi_b \in \mathcal{F}(X)$ iff $\phi’ \circ p_1=\phi’ \circ p_2$, where the $p_i$ are the projections $X_K \times_X X_K \rightarrow X$. It is then easy to see that such a $\phi_b$ is a morphism whose base change to $K$ is $\overline{\phi}$, and thus extends $\phi$.
Let $f_i=\phi’ \circ p_i$, and we want to show that $f_1=f_2$. Let $U \subset X$ be an open subset such that $\phi$ is defined on $U$. Then $\overline{\phi}_{|U_K}$ is a base change of $\phi$, so that $f_1$ and $f_2$ have the same restriction to $U_K \times_U U_K$. If $Z \in \{U,X\}$, $Z_K \times_Z Z_K \cong Z \times_k (K \otimes_k K)$ (functorially in $Z$), and the spectrum of $K \otimes_k K$ is a finite set of isolated points, so that $U_K \times_U U_K$ is an everywhere dense open subset of $X_K \times_X X_K$.
As $Y_K$ is separated over $k$, it follows that $f_1=f_2$ topologically and thus that, given any local section $s$ of $Y$, $f_1^{\sharp}s -f_2^{\sharp}s$ is a nilpotent section vanishing on $U_K \times_U U_K$.
To conclude, it is thus enough to show that if $A$ is a geometrically integral $k$-algebra (the ring of sections of any open affine subset $V$ of $X$) and $B$ is a localization of it (the ring of sections of some principal open subset of $V \cap U$), then the base change $A \otimes_k C \rightarrow B \otimes_k C$ is injective (where $C=K \otimes_k K$). But this is true because $C$ is flat over $k$.
Best Answer
The closed points of a finite type $k$-scheme are precisely the points with residue extension $k(x)/k$ algebraic (equivalently finite). The residue field of a closed point is a domain that is finitely generated as a $k$-algebra, also a field, hence a finite extension of $k$ by (a form of) the Nullstellensatz. For the other implication, assume $X=\mathrm{Spec}(A)$ with $A$ a finitely generated $k$-algebra. If $\mathfrak{p}\in X$ is such that $k(\mathfrak{p})/k$ is algebraic, then $k(\mathfrak{p})$ is integral over the domain $A/\mathfrak{p}$ which is therefore a field, i.e., $\mathfrak{p}$ is maximal.