Vector space basis theorem, can it be generalized to R-modules?

Question

Observe the following theorem for vector spaces.

Let $V$ be a finite-dimensional, nontrivial vector space, and suppose $V=\operatorname{span}(v_1,v_2)$. Then some subset of $\{v_1,v_2\}$ will be a basis for $V$.

I would like to know if there is a corresponding theorem for R-modules (where $R$ is a commutative ring with 1), or if not, at least for free R-modules. Stating such a theorem would be useful for me, but attaching a simple proof to accompany it would be even better.

Edit

Just to give some context, I want to work with the ring $R=\mathbb{Z}[\sqrt{-5}]$. So a corresponding theorem which will work for at least this ring (even if it doesn't work for all rings) should be good enough for me.

Answer 1

No: take $V = \Bbb{Z}$ viewed as a $\Bbb{Z}$-module. $V$ is spanned by any coprime pair of integers $(v_1, v_2)$, e.g., $v_1 = 2$ and $v_2 = 3$, but it only has two bases: $\{1\}$ and $\{-1\}$.

Answer 2

The statement is wrong for general $R$-modules, even for free ones: Consider $\mathbb Z$ as a $\mathbb Z$-module, then $\mathbb Z=\langle 2,3\rangle$, but neither $2$ nor $3$ generates $\mathbb Z$, hence $\{2,3\}$ does not contain a basis even though $\mathbb Z$ is free.