I am probably making a really silly mistake but I can't figure it out:
Let $\mathbb F_q$ be a finite field and $a$ an element in it that is not a square. Let $E$ be the elliptic curve corresponding to the equation $y^2 = x^3 + 1$ and $E'$ it's twist $ay^2 = x^3 + 1$. Let $f,g: E, E' \to \mathbb P^1$ be the projection to the $x$ co-ordinate and let $C = E\times_{\mathbb P^1} E'$.
Now, $C$ is a possibly singular curve and let $\pi: \tilde C \to C$ be it's normalization. This is a birational map away from finitely many points and therefore $|\tilde C(\mathbb F_{q^r})| - |C(\mathbb F_{q^r})|$ is bounded independent of $r$. By the Weil bound, $|\tilde C(\mathbb F_{q^r})|$ (and hence $|C(\mathbb F_{q^r})|$) goes to infinity as $r \to \infty$.
On the other hand, a $\mathbb F_{q^r}$ point (with $r$ odd) of $C$ corresponds to a point each on $E,E'$ such that their $x$ co ordinates are equal. But if we fix a value for $x$, then $x^3+ 1$ is either a square in $\mathbb F_{q^r} $ or it isn't and this determines whether there is a corresponding point on $E$ or on $E'$ since r is odd and so the twists continue to be non isomorphic.
In particular, there are no pairs of points on $E,E'$ with the same $x$ co ordinate (apart from finitely many points at infinity independent of $r$) and therefore only finitely many $\mathbb F_{q^r}$ points on $C$, independent of $r$.
I have reached a contradiction so what mistaken assumption did I make?
I think the point may be that in the Weil bound we require $\tilde{C}$ to be absolutely irreducible (or geometrically irreducible)? If we don't, we could have examples like $\mathbb{P}_{\mathbb{F}_{q^{2}}}^{1}$ which is smooth projective over $\mathbb{F}_{q}$ but has an $\mathbb{F}_{q^{r}}$-point (if and) only if $r$ is even. In your case, an affine chart of $C$ is isomorphic to $\operatorname{Spec} \mathbb{F}_{q}[x,y_{1},y_{2}]/(y_{1}^{2}=x^{3}+1 , ay_{2}^{2}=x^{3}+1)$ which is not geometrically irreducible because of the relation $y_{1}^{2} = ay_{2}^{2}$; thus the normalization $\tilde{C}$ is not geometrically irreducible either.