Find all $z$ such that $$(z+1)^6 = z^6.$$
I have moved $z + 1$ to the other side such that $$\left(\frac{z}{z+1}\right)^6 = 1,$$ which then makes it so
$$\frac{z}{z+1} = 1 = \frac{z}{z+1} = e^{ik\pi/3}.$$
I know $k \neq 0$ where $k = 0, 1, 2,\dotsc ,5$, but I don't know where to go from here.
You have already found that $$\frac{z}{z+1}=e^{ik\frac{\pi}{3}}$$
This can be rearranged to get $$z=\frac{e^{ik\frac{\pi}{3}}}{1-e^{ik\frac{\pi}{3}}}$$
You can simplify this using trig form and double angle identites to get your solution as $$z=-\frac 12+i\cot \left(k\frac{\pi}{6}\right)$$