In fact, $y^2 = x^3 - x + 1$ is an elliptic curve. I have to show that the number of $\mathbb{F}_p$-rational points on this curve is at most $2p$.
What I know is that for an elliptic curve $E : y^2 = x^3 + Ax + B$, $E(\mathbb{F}_p)$ has at most $2p+1$ points. By Hasse's Theorem $\mathrm{Card}(E(\mathbb{F}_p)) = p+1-a_p(E)$ with $\left| a_p(E) \right| \leq 2\sqrt{p}$. But I am not sure if this could help me to solve my problem. Moreover, consider the set $S= \{(x, y) \in \mathbb{F}_p^2 \mid y^2 = x^3\}$. How can I show that $\mathrm{Card}(S) = p$ for any prime $p$ ?
Are you including the point at infinity? If not then this is easy: each $x$ value determines at most two $y$ values. If you are, then Hasse is certainly strong enough. But for an elementary proof... all the values of $x^3-x+1$ would have to be quadratic residues for a counter-example, but I don't see any down-to-earth reason why this can't be true.
By the way, $y^2=x^3$ is not an elliptic curve. It is parametrised by $(x,y)=(t^2,t^3)$ though.