I agree with Charles Matthews, there's something that doesn't seem right in the formulation of the problem. Consider what you get for $m=2$:
$$\begin{align} {2 \choose 1} &= R_1 + C_1 \\\ {2 \choose 2} &= R_2 + R_1C_1 + C_2\\\ {2 \choose 3} &= R_3 + R_2C_1 + R_1C_2 + C_3\\\ {2 \choose 4} &= R_4 + R_3C_1 + R_2C_2 + R_1C_3 + C_4 \end{align}$$
Given the assumptions made for $k>m$, this becomes
$$\begin{align} 2 &= R_1+C_1\\\ 1 &= R_2 + R_1C_1 + C_2\\\ 0 &= R_2C_1 + R_1C_2\\\ 0 &= R_2C_2 \end{align}$$
This has just three possible solutions $(R_1,C_1,R_2,C_2)$: $(0,2,0,1)$ $(1,1,0,0)$, and $(2,0,1,0)$, so it's not clear what's meant by saying "Given $R_1$, the system has a unique solution." Here's my guess:
For each $m$, the system has $m+1$ solutions, each solution is in non-negative integers, and there is exactly one solution for each integer $0 \le R_1 \le m$.
This is certainly true for $m=1$ and $m=2$ and seems to be true for $m=3$ as well (I checked but didn't doublecheck that case).
Added later: Aaron Meyerowitz has taken this quite a bit further, so read his answer, not this one.