What are the irreducible representations of $(\mathbb{Z},+)$?

You are probably looking for the complex representations (is that right?). Firstly I claim any finite-dimensional simple rep is one-dimensional. Let $G=\langle g\rangle$ be infinite cyclic, this will stand in place of $\mathbb{Z}$. If $g$ acts on some nonzero finite-dimensional complex vector space $V$ by a linear map $\rho(g)$ then this linear map has an eigenvector, which spans a one-dimensional subrep. If the rep was irreducible, this subrep must be all of $V$, so $\dim V=1$.

Now note that if $\lambda \in \mathbb{C}^*$ (I mean the non-zero complex numbers) then $g \mapsto \lambda$ induces a homomorphism $\rho_\lambda: G \to \mathsf{GL}_1(\mathbb{C})$, i.e. a representation of $G$. Clearly any one-dimensional representation arises this way. Furthermore if $\lambda \neq \mu$ then the representations afforded by $\rho_\lambda$ and $\rho_\mu$ are not isomorphic: not much conjugacy happens in $\mathsf{GL}_1$ !

In conclusion, the simple representations are in natural bijection with $\mathbb{C}^*$: this is called the character group.

If $\rho\colon \mathbb Z\to\mathrm{GL}(n,\mathbb C)$ is a complex representation, then because $\rho$ is a homomorphism, it is entirely determined by $\rho(1)$ (since $\rho(n)=\rho(1)^n$). So if $\rho$ and $\rho'$ are two non-isomorphic representations of $\mathbb Z$, then $\rho(1)$ and $\rho'(1)$ determine two non-conjugate matrices in $\mathrm{GL}(n,\mathbb C)$. Since two matrices are conjugate over $\mathbb C$ iff they have the same Jordan canonical form, we get that the $n$-dimensional complex representations of $\mathbb Z$ (up to isomorphism) are in one-to-one correspondence with the $n\times n$ Jordan canonical forms.

EDIT: I didn't notice you asked for irreducible representations of $\mathbb Z$. In this case, I guess see mt_'s answer.