Show these polynomial rings are not isomorphic

$\mathbb{R}[T, T^{-1}]$ is the localization of a UFD (a PID, even), hence is a UFD. However, $\mathbb{R}[X,Y]/(X^2 + Y^2 - 1)$ is not a UFD: the idea is that $X^2 = (1-Y)(1+Y)$ gives two different factorizations. For details, see this answer.

The interesting part is that if we work over $\mathbb{C}$, the two corresponding rings $\mathbb{C}[X,Y]/(X^2 + Y^2 - 1)$ and $\mathbb{C}[T,T^{-1}]$ are isomorphic: $$ \mathbb{C}[T,T^{-1}] \cong \mathbb{C}[T,U]/(TU - 1) \cong \mathbb{C}[X,Y]/(X^2 + Y^2 - 1) $$ where the last isomorphism sends $T \mapsto X + iY$ and $U \mapsto X - iY$.

If $\xi$ and $\eta$ are the images of $X$ and $Y$ in the quotient ring, then $\xi^2+\eta^2=1$, so you should find two such elements in $\mathbb{R}[T,T^{-1}]$.

An nonzero element in $\mathbb{R}[T,T^{-1}]$ can be uniquely written as $T^mf(T)$, where $m$ is an integer and $f(T)\in\mathbb{R}[T]$ with $f(0)\ne0$. So suppose $$ (T^mf(T))^2+(T^ng(T))^2=1 $$ Suppose $m\ge0$ and $n\ge0$. Then the leading coefficient of the left-hand side is positive, so the degree has to be zero, which implies $m=n=0$ and $f$ and $g$ constant.

However, the images of $\xi$ and $\eta$ should generate $\mathbb{R}[T,T^{-1}]$: contradiction.

Suppose $m<0$ and, without loss of generality, $m\le n$. Then we get $$ f(T)^2+T^{2n-2m}g(T)^2=T^{-2m} $$ Evaluating at $0$, we obtain $f(0)^2=0$ or $f(0)^2+g(0)^2=0$ (according to $n\ne m$ or $n=m$): in both cases it is a contradiction.