Why is $\mathbb{Z}$ not a Kähler group?

If $X$ is a compact Kähler manifold, then $h^{p,q}(X) = h^{q,p}(X)$ and $b_k(X) = \sum_{p+q=k}h^{p,q}(X)$, so in particular, $b_1(X) = h^{1,0}(X) + h^{0,1}(X) = 2h^{1,0}(X)$ is even. Now,

$$b_1(X) = \operatorname{rank} H^1(X; \mathbb{Z}) = \operatorname{rank} \operatorname{Hom}(\pi_1(X), \mathbb{Z}).$$

If $\pi_1(X) = \mathbb{Z}$, $b_1(X) = 1$ which is not even, so $X$ is not Kähler. More generally, if $\pi_1(X)$ is abelian, then it must have even rank.

Note, there are non-compact Kähler manifolds with fundamental group $\mathbb{Z}$, for example $X = \mathbb{C}^*$.

Every free group is not Kähler (fundamental group of compact Kähler manifold). In particular, a free group of rank 1 is not Kähler.

This follows since the rank of the abelianization of a Kähler group is even, even rank free groups contain finite index subgroups of odd rank (see here for example), and finite index subgroups correspond to covers of Kähler manifolds which are also Kähler.

This and other interesting examples can be found in Fundamental Groups of Compact Kähler Manifold by Amorós, Burger, Corlette, Kotschick, and Toledo.