If you are considering consider rings that are not necessarily commutative, here's an example: let $V$ be a countable dimensional vector space over a field $F$, and let $A$ be the ring of all endomorphisms of $A$. I claim that $A\cong A\oplus A$ (as left $A$-modules); if so, then using (and iterating) this isomorphism you can find maximal linearly independent subsets of any finite cardinality.
To see that $A\cong A\oplus A$, it suffices to exhibit a two-element $A$-basis for $A$. Let $e_1,e_2,\ldots$ be a basis for $V$. Let $f_1\in A$ be the endomorphism that maps $e_2,e_4,e_6,\ldots$ to $e_1,e_2,e_3,\ldots$, respectively, and maps every odd-indexed basis element to $0$; let $f_2\in A$ be the endomorphism that maps $e_1,e_3,e_5,\ldots$ to $e_1,e_2,e_3,\ldots$, and maps the even-indexed basis elements to $0$. Then $f_1,f_2$ spans $A$: if $\varphi \in A$, then we can write $\varphi$ as $\varphi=gf_1+hf_2$, where $g(e_i)=\varphi(e_{2i})$ and $h(e_j)=\varphi(e_{2j-1})$. To see that $f_1$ and $f_2$ are $A$-linearly independent, suppose that $af_1+bf_2=0$; evaluating at the odd indexed $e_i$ shows that $b(e_j)=0$ for all $j$, and evaluating at the even indexed $e_i$ shows $a(e_j)=0$ for all $j$. Thus, $f_1,f_2$ is also a basis for $A$, which gives an isomorphism $A\cong A\oplus A$. Being bases, they are certainly maximal linearly independent sets.
