From this answer, I can see that to show that for the $2$-dimensional case, I can use
$$ t\mapsto\begin{cases} (x,y)=\left( \dfrac{1-t^2}{1+t^2}, \dfrac{2t}{1+t^2} \right) & \text{if }t\ne\infty, \\[10pt] (x,y)=\lim\limits_{t\to\infty} \left( \dfrac{1-t^2}{1+t^2}, \dfrac{2t}{1+t^2} \right) & \text{otherwise}. \end{cases} $$
But what about the $3$-dimensional case? I feel like it should look something like this:
$$ t\mapsto\begin{cases} (x,y,z)=\left( \dfrac{1-t^2}{1+t^2}, \dfrac{2t}{1+t^2}, t \right) & \text{if }t\ne\infty, \\[10pt] (x,y,z)=\lim\limits_{t\to\infty} \left( \dfrac{1-t^2}{1+t^2}, \dfrac{2t}{1+t^2},t \right) & \text{otherwise}. \end{cases} $$
as that's kind of what you do when you parametrize a sphere, but it doesn't work. We have that $t=\dfrac{y}{x+1}$, so at $z=0$ we would have $y=0$, which wouldn't describe a circle at all.