so it suffices to show that every non-trivial finite field extension of $\Bbb R$ has degree 2. Let $K / \Bbb R$ be a finite nontrivial field extension. Then $K ^\times$ is a connected, abelian Lie group, so $K^\times \cong \Bbb R^d \oplus (S^1)^k$ of some $d$ and $k$ such that $d+k=[K:\Bbb R]$. Note that $k\geq 1$, since $K^\times$ has torsion, e.g. $-1$. Thus $\pi_1(K^\times) = (\Bbb Z)^k \neq 0$.
But $K^\times = \Bbb R^n \setminus \{0\}$, so the only way that this is not simply connected is $n=2$

Write $k[X,Y]/(X-XY^2)=R$ and by abuse of notation let $x$ and $y$ be the residue classes of $X$ and $Y$. Suppose $xy=ux$, for a unit $u$ with inverse $v$. Let $u$ and $v$ be the residue classes of $U$ and $V$ in $k[X,Y]$, then $UV-1=(X-XY^2)A$ and $UX-XY=(X-XY^2)B$. The second equation implies $U-Y=(1-Y^2)B$. Now set $X=0$, so that we work in $k[X,Y]/(X) = k[Y]$.
Then the equation $UV-1=(X-XY^2)A$ implies $U(0,Y)V(0,Y)=1$, so that $U(0,Y) \in k^\times$. But from the equation $U-Y=(1-Y^2)B$, we get $U(0,Y)=(1-Y^2)B(0,Y)+Y$. The RHS is certainly not constant in $Y$