We may assume, by replacing $G$ by the Zariski closure of $G(k)$, that $G(k)$ is Zariski dense. By assumption, $G(k)$ is $\mathbb Z$; hence $G$ is abelian and after replacing $G$ by the connected component of identity of $G$, we may assume that $G$ is connected. \vskip 5mm

Now a conneced abelian algebraic group is a product of ${\mathbb G}_a$ and ${\mathbb G}_m$. Hence the assumptions mean that $k$ or $k^*$ must be $\mathbb Z$; this is impossible for a subfield of $\mathbb C$.

In detail, $G$ is a product of copies of ${\mathbb G}_a$ and copies of ${\mathbb G}_m$. Hence $G(K)$ is the product of (copies of) $K$ and $K^*$. Assume that a factor like a product of ${\mathbb G}_a$ occurs. Then $G(k)$ contains the subgroup $k\supset {\mathbb Q}$. However, $G(k)$ being $\mathbb Z$, any infinite subgroup must be $\mathbb Z$ but $\mathbb Q$ is not $\mathbb Z$.

A similar but more involved reasoning may be used to tackle the case when $G$ is a product of the multiplicative group ${\mathbb G}_m$ over $K$.

Since the questioner has asked for details, if we assume that $G$ over $K$ does not have the additive group components, then over the smaller field $k$, $G$ is isogenous to a product of ${\mathbb G}_m$ and the groups $R^1_{l/k}({\mathbb G}_m$ where $R_1$ refers to the group of norm one elements. This is a theorem due to Ono (for reference, one may see Borel-Tits). The ${\mathbb G}_m$ factor may easily be taken care of. As to the other group $R^1$, one has to work a little to say that the group of norm one elements in $l$ cannot be isomorphic to $\mathbb Z$, for any char zero $l/k$.