Let $G$ be an affine algebraic group defined over a field of characteristic zero $K$. Suppose $G$ has only one single $K$point, can we conclude that $G$ does not have more points?

Question (edited here)
As shown by David Speyer in the comments, if $\dim G=0$ then yes. For example, let $G$ be the solutions to $z^31$. Then over $G(\mathbb{Q})=1$, but $G(\mathbb{C})=3$ and hence $G$ is not trivial. On the other hand, the comments by Brian Conrad show that if $\dim G \geq 1$, then $G(k)\not=1$. I think this proves it: Since the identity component of $G$ is a connected affine algebraic group over $k,$ it suffices to prove this for $G$ connected. Then, since we are in characteristic 0, $G$ is isomorphic (as a variety, but not as an algebraic group) to $(G/G_u) \times G_u$ where $G_u$ is its unipotent radical. The unipotent radical is likewise isomorphic to an affine space, and $G/G_u$ is reductive. By the Bruhatdecomposition $G/G_u$ contains an affine open subset whose $\overline{k}$points are isomorphic to $(\overline{k}^*)^n \times \overline{k}^m$ where $\overline{k}$ is an algebraic closure of $k$. 

