Recall that a ring $R$ is called right (left) Noetherian if every right (left) ideal $I$ of $R$ is a finitely generated $R$-module, i.e., there exists $x_1,\ldots,x_m \in I$ such that $I=x_1R+\ldots+x_mR$ (or $I=Rx_1+\ldots+Rx_m$).
Suppose that $R$ is finitely generated as an additive group, i.e., $R=\mathbb{Z}x_1+\ldots+\mathbb{Z}x_m$ for some $x_1,\ldots,x_m \in R$. Is it true that every right (or left) ideal of $R$ is finitely generated as an $R$-module?
Yes, it is true.
The ring $\mathbb{Z} $ is a PID, hence noetherian. Thus an abelian group (i.e. $\mathbb{Z} $-module) is a noetherian $\mathbb{Z} $-module iff is finitely generated. Now, given any ring homomorphism $S\to R$ and an $R$-module $M$, if $M$ is noetherian as an $S$-module, so it is as an $R$-module. Taking $S=\mathbb{Z} $ and the unique ring homomorphism $\mathbb{Z} \to R$, your assertion follows.
Every right ideal of $R$ is in particular a $\mathbb Z$ submodule. The hypothesis says $R_\mathbb Z$ is a finitely generated module over a Noetherian ring ($\mathbb Z$), hence it is a Noetherian module. So any ascending chain of right ideals you pick from $R$ is a chain of $\mathbb Z$ submodules in a Noetherian module, so it must terminate.
