As observed already, you don't have there an equation of the form $P(z)=0$ for some polynomial $P$, but rather an equation of the form
where $P$ is a polynomial in two variables of degree, say, $d$. After the substitutions $z=x+iy$ and $\bar z=x-iy$, and separating the real from the imaginary part, equation $(\star)$ is transformed into a system
where $P_1$ and $P_2$ are polynomial with real coefficients and you are looking for the real solutions of $(\star\star)$.
Since $\deg(P_1)=\deg(P_2)=\deg(P)=d$, if $(\star\star)$ does not admit infinitely many solutions (which may as well happen, see the comment of Marvis) you may expect up to $d^2$ solutions by Bezout's Theorem.