Inspired by this question which in the case $k=2$ we would need to factor the second factor too:
$$\displaystyle\sum_{l=0}^{n-1} m^l$$ Can we say anything in general for which $n$ or $m$ this will be possible?
My own work is limited to concluding that if $n$ is a composite number $n = f_1\cdot f_2$, then:
$$\displaystyle\sum_{l=0}^{n-1} m^l = \left(\displaystyle\sum_{l=0}^{f_1-1} m^l\right)\left(\displaystyle\sum_{l=0}^{f_2-1} m^{l\cdot f_1}\right)$$
For example $n=6 = 3 \cdot 2$ : $$m^5+m^4+m^3+m^2+m^1+1 = (m^2+m^1+1)(m^3+1) = (m+1)(m^4+m^2+1)$$
If $s(m, n) =\sum_{l=0}^{n-1} m^l $, then $s(m, n) =\dfrac{m^n-1}{m-1} $.
Also, from $x^n-1 =(x-1)\sum_{j=0}^{n-1} x^j$, we have the factorizations $m^{ab}-1 =(m^a-1)\sum_{j=0}^{b-1} m^{ja} =(m^a-1)s(m^a, b) $ and $m^{ab}-1 =(m^b-1)\sum_{j=0}^{a-1} m^{jb} =(m^b-1)s(m^b, a) $.
Therefore, if $n = ab$ then $s(m, n) =\dfrac{m^n-1}{m-1} =\dfrac{m^{ab}-1}{m-1} =\dfrac{(m^a-1)s(m^a, b)}{m-1} =s(m, a)s(m^a, b) $.
We also have $s(m, n) =s(m, b)s(m^b, a) $.