Group over algebraic curves having genus greater than 1

Let $G/k$ be a proper smooth connected non-trivial group variety. For $1 \neq x \in G$, the translation by $x$ has no fixed point, so by the Lefschetz fixed point formula for $\ell$-adic cohomology (using $G$ connected), the Euler characteristic of $G$ is $0$. If $G/k$ is $1$-dimensional, this implies that $0 = \chi(G) = 2-2g$, so $g=1$.

The following simple proof works as soon as the curve contains infinitely many points (for instance, when it is defined over an algebraically closed field).

If you have an algebraic group $G$, then for every fixed $a \in G$ the translation $x \mapsto x+a$ is an automorphism of the underlying algebraic variety $X_G$, in particular $G$ embeds into $\mathrm{Aut}(X_G)$.

But it is well known that a curve of genus $g \geq 2$ has at most finitely many automorphisms (for instance, in characteristic $0$ there are at most $84(g-1)$ of them).

You seem to be interested in group structures on complete (= proper) curves. In general, if you have a complete group variety X of dimension $d$ over $\mathbb{Q}$, then you can prove that the tangent (or cotangent) sheaf is free. So you get an isomorphism $\Omega_X \cong \mathcal{O}_X^{\oplus d}$. This means that $\dim H^0(X,\Omega_X) = d \cdot \dim H^0(X,\mathcal{O}_X) = d$.

If $X$ is a curve, so $d = 1$, note that $\dim H^0(X,\Omega_X)$ is the genus of $X$. Hence $X$ has genus $1$.

(Fun fact: Since $X$ is proper, one can also show that the group structure must be commutative!)