Let $f \in \mathbb{Q}[x,y]$ be a polynomial, define a curve $C = \{ f(x,y) = 0 \}$. Suppose the genus of $C$ is $0$ and $C(\mathbb{Q}) \neq \emptyset$. Then $C(\mathbb{Q})$ is birational to $\mathbb{A}^1_\mathbb{Q}$.
I have heard of this result, but I have not been able to find a reference; all the sources I have looked so far deals with algebraically closed fields. Could someone please suggest me a reference? Thank you.
This is the most well-known example of Hasse (local-global) principle, which does not hold in higher genus. In your particular case, it is easy. Basically the proof is the same as finding all the rational solutions to $x^2+y^2=1$ as the following:
Assume $C$ is smooth. Then the genus formula says $f$ is of degree [at most] $2$. Use rational slope lines passing through the rational points to intersect the curve $f=0$, namely, solve equations in terms of the slope. Plugging in the linear curve into the quadratic equation. It factors into two solutions, one is the given rational point, the other gives you a parametrization.
If $C$ is not smooth, the degree might be higher. For example, $y^2=x^3-x^2$ is a rational curve. But you can still use this method, since the genus $0$ assumption basically allows you to cancel terms.
[Edit]: As reuns remarked, some extra assumptions are needed for the statement to be correct. One way to fix this is by restricting to the smooth case (apparently my careless writing skips that point):
If $C\subseteq \mathbb{A}^2$ is a smooth curve of genus $0$ defined over $\mathbb{Q}$ with a $\mathbb{Q}$-point, then $C$ is rational.