This question has been asked before at the link above, but I am trying to prove it in a somewhat different vein, largely because I spent a while on it before looking for alternative answers and can't get past a certain point. Maybe some of my intermediate steps are incorrect.
Let $v\in R$, $v\neq0$. Consider the cyclic $R$-module $\langle\langle v \rangle\rangle = \{rv\mid\ v\in R\}$. This is a finitely generated $R$-module and so has a basis by assumption, specifically $\{v\}$. Then $\{v\}$ is linearly independent, and so for any $r\in R$, $rv=0$ implies that $r=0$. Since $v$ is an arbitrary non-zero element in $R$, we can apply the same method to any other non-zero element $u$ of $R$ and conclude that for any nonzero $u \in R$, $ru=0$ implies $r = 0$. So $R$ has no zero divisors and is thus an Integral Domain.
First of all, are the assertions I have made thus far correct? If not, please let me know. If this is correct to this point, is there a way to extend this to demonstrate the existence of multiplicative inverses in $R$ and thus to show that $R$ is a field in a way that does not rely on the $R/I$ approach used in the link above?