I found exactly the same counterexample as DHMO. However, I want to share how I found it, so maybe you can come up with many more counterexamples.
I started setting $z_1=re^{i\theta},z_2=\rho e^{i\phi}$. The hypothesis then becomes:
$$re^{i\theta}(1+\rho)=i\rho e^{i\phi}(1+r).$$
Taking moduli, this yields:
$$r(1+\rho)=\rho(1+r)\iff r=\rho.$$
So $|z_1|=|z_2|$. Then I took arguments:
$$\theta=\phi+\frac{\pi}{2}+2k\pi\iff\theta-\phi=\frac{\pi}{2}+2k\pi.$$
Since this approach didn't seem to be leading me to a proof, I tried a disproof. Since the moduli are equal, I thought, why not set them to 1, simplifying the equations? The hypothesis, with moduli equal to 1, becomes:
$$2z_1=2iz_2\iff z_1=iz_2\iff z_1=e^{i\theta},z_2=e^{i(\theta-\frac{\pi}{2})}.$$
Plug into thesis:
$$0=z_1+\overline{z_2}=e^{i\theta}+e^{i(\frac{\pi}{2}-\theta)}=e^{i\theta}+ie^{-i\theta}=z_1+i\overline{z_1}.$$
Let's switch to algebraic form: $z_1=a+ib$. This gives, plugging into thesis:
$$0=a+ib+i(a-ib)=a+b+i(a+b)\iff a=-b.$$
So $z_1$ should be a real multiple of $1-i$, which with modulus 1 implies:
$$z_1=\frac{1}{\sqrt2}(1-i).$$
So just take any other $z_1$ with modulus one, and you will find a counterexample. For example, $z_1=i$. Plug into hypothesis to find $z_2$, always under the assumption that $|z_2|=|z_1|=1$:
$$0=z_1(1+|z_2|)-iz_2(1+|z_1|)=2i-2iz_2\iff z_2=1.$$
Plug into thesis, it doesn't work. Many more counterexamples can be found.
Note that, since the moduli are equal, you can just assume $|z_1|=r$ for any $r$, and the thesis and hypothesis merely get multiplied by $r^2$, so you still get $z_1$ is a multiple of $1-i$, which is very restrictive.