If some ideal is not finitely generated, there is one maximal among the ideals that are not

Inclusion partially orders $\mathrm{\Sigma }$, and $\mathrm{\Sigma }$ is nonempty by hypothesis. The empty chain is bounded in $\mathrm{\Sigma }$: its upper bounds are all the elements of $\mathrm{\Sigma }$, and there is at least one.

Let $\mathcal{C}\subseteq \mathrm{\Sigma }$ be a nonempty chain and put $U:=\bigcup \mathcal{C}$. Then $U$ is an ideal of $R$. It is nonempty, since some $I\in \mathcal{C}$ contains $0$. For $x,y\in U$ pick $I,J\in \mathcal{C}$ with $x\in I$ and $y\in J$; the two are comparable, so both lie in the larger one, whose being an ideal gives $x-y$ there and hence in $U$. For $r\in R$ and $x\in U$, choosing $I\in \mathcal{C}$ with $x\in I$ gives $rx\in I\subseteq U$. The ideal criterion applies.

$U$ is not finitely generated, so $U\in \mathrm{\Sigma }$ and $U$ is an upper bound of $\mathcal{C}$. Suppose instead $U=(u_1,\dots ,u_k)$ with $k\in \mathbb{N}$; each $u_i$ lies in $U$, hence in some member of $\mathcal{C}$. A nonempty finite subset of a chain has a greatest member, by induction on its size using comparability of any two elements, so there is $I\in \mathcal{C}$ containing every $u_i$; when $k=0$ take any $I\in \mathcal{C}$, which exists because $\mathcal{C}$ is nonempty. Then $U=(u_1,\dots ,u_k)\subseteq I\subseteq U$, so $I=U$ is finitely generated, contradicting $I\in \mathrm{\Sigma }$.

Every chain in $\mathrm{\Sigma }$, empty or not, therefore has an upper bound in $\mathrm{\Sigma }$, and $\mathrm{\Sigma }$ is nonempty; Zorn's lemma gives a maximal element of $\mathrm{\Sigma }$. This is the one place the argument leaves ZF, and it uses the full axiom of choice rather than a countable or dependent form.