Find all complex numbers z such that $(z+1)^4 = (z+i)^2$ I have the following problem:

Find all complex numbers z such that $(z+1)^4 = (z+i)^2$

I'm stuck after trying to see it as a difference of squares:
$((z+1)^2)^2-(z+i)^2=0$
$((z+1)^2 + (z+i)) ((z+1)^2-(z+i))=0$
I've tried expanding it and comparing it to a binomial form (such as $a+bi$). But I end up with something that doesn't seem right and I still can't come to an answer.
Any suggestions will be much appreciated.
Thanks

Note that $$\left(\frac{(z+1)^2}{z+i}\right)^2=1=e^{2\pi i n}$$
for $n\in\mathbb Z$. So then,
$$\frac{(z+1)^2}{z+i}=e^{\pi i n}=(-1)^n$$
Now we just need to solve this quadratic properly.
