The answer to your second question is $R=2$.

Suppose a zero-divisor $a$ has rank 2 in a torsion-free group G.

W.l.o.g. there is a $b$ such that $a\cdot b = 0$.
By multiplying with a suitable group element from the left we can achieve that $a = 1 + g$ for some $1\neq g \in G$.
Similarly by multiplication from the right we can achieve that $b$ has the form $b= 1 + h_2 + \ldots + h_{k}$, where $k$ is the rank of $b$, and all $h_i$ are distinct and different from 1. For notational simplicity we define $h_1 = 1$.

Since $a \cdot b = 1 + h_2 + \ldots + h_{k} + g + g h_2 + \ldots + g h_{k} = 0$, and since $g h_i \neq g h_j$ for $i\neq j$ there is a matching that pairs every element in $A = \lbrace 1, h_2, \ldots, h_{k} \rbrace$ with an element in $B= \lbrace g, g h_2, \ldots, g h_{k}\rbrace$. The elements that are paired are equal (i.e., $h_i = g h_j$ if $h_i$ is paired with $g h_j$).

From the matching we conclude that there is an index $i_1$, such that $g = h_{i_1}$.

Then there is an index $i_2$ (different from indices previously used) such that $g^2 = g h_{i_1} = h_{i_2}$.

Then there is an index $i_3$ (different from indices previously used) such that $g^3 = g h_{i_2} = h_{i_3}$.

And so on $\ldots$

By induction, some index $i_t$ must be equal to 1 and $g^t = h_1 = 1$ must hold for some $t\in\mathbb{N}$.
Which shows that G has torsion and yields a contradiction.

For all I know, a similar statement is not known for $R = 3$. However, the Conjecture itself over $\mathbb{Q}$ is equivalent to $R = \infty$. I have been researching this question for $R= 3$, and the proof does not seem to adapt easily. However it is possible to show statements of the following form: If $a\cdot b = 0$ then $a$ must have rank larger than $R_1\in \mathbb{N}$ or $b$ must have rank larger than $R_2\in \mathbb{N}$.
Such a statement can be reduced to a finite case analysis (potentially involving undecidable torsion-freeness questions), which is still doable by hand for $R_1=4$ and $R_2 = 4$. However, the number of cases in the reduction (of the finite case analysis I know) grows like the double factorial.