Stuck on a system of equations

I got the following system of equations but I get stuck after a couple of steps.

$$1+x+2zy = 0$$ $$1+y+2zx = 0$$ $$y^2+x^2-1 = 0$$

From the third equation, I get:

$$x = \pm \sqrt{1-y^2}$$ and $$y = \pm \sqrt{1-x^2}$$

From the first two equations I get the following expression:

$$x =\frac{-1-y}{2z}, y =\frac{-1-x}{2z} $$

But then what? I just don't see how I can get all possible expressions for $z, y$ and $x$ where all equations hold.

I'd appreciate an hint on how to move on with this.

From the first and second equations, you can get $$\frac{y}{x}=\frac{1+x}{1+y}$$ or $$y^2+y=x^2+x$$ Adding 1/4 to both sides allows you to complete the square, yielding $$y+1/2 = \pm(x+1/2)$$ so $$y=x,-x-1$$ You can substitute these into the third equation (Note: there might be extraneous solutions.)

The last equation defines a circle of radius 1 (or a cylinder, in $xyz$ coordinates.

Substitute: $x=r\cos t$,$y=r\sin t$ to obtain: $$\begin{cases} 1+r\cos t+2zr\sin t=0\\ 1+r\sin t +2zr\cos t=0\\ r^2=1 \end{cases}$$ Subtract the 1st equation from the 2nd, $r$ is eliminated: $$\begin{cases} 2z(\cos t-\sin t)=\cos t-\sin t\\ r^2=1 \end{cases}\Rightarrow z=\frac{1}{2}$$ Now go back to the original system of equations. You'll get a circle and a line that will intersect.

Edit: the other solutions may be obtained from the last systems of equations as well, if you notice that both equations are correct when $\sin t=\cos t$. It only occurs when $\sin t=\cos t=\pm\frac{1}{\sqrt{2}}$, and from here you can also find $z$.

Following Paul's approach, all solutions over $\Bbb C$ are in fact real, and given by $$ (x,y,z)=\left(-1,0,\frac{1}{2}\right), \left(0,-1,\frac{1}{2}\right), \left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}, \frac{- \sqrt{2} -2}{2\sqrt{2}}\right), \left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}, \frac{2 - \sqrt{2}}{2\sqrt{2}}\right) $$