$(y+zq)^2=z^2(1+p^2+q^2)$.
How to find the complete integral of the given pde by using Charpit's equations ?
Charpit's equations are:
$$ \frac{dx}{2pz^2}=\frac{dy}{2qz^2-2z(y+zq)}=\frac{dz}{2p^2z^2-2qz(y+zq)}=\frac{dp}{-2pz(1+p^2+q^2)}=\frac{dq}{2(y+zq)-2qz(1+p^2+q^2)} $$
Let $u:=z^2$, so that $z=\pm\sqrt{u}$ and $(p,q)=(z_x,z_y)=\left(\pm\frac{u_x}{2\sqrt{u}},\pm\frac{u_y}{2\sqrt{u}}\right)=:\left(\frac{P}{2z},\frac{Q}{2z}\right)$. In terms of the new variables $u, P,$ and $Q$, the PDE reads $$ \left(y+\frac{Q}{2}\right)^2=u+\frac{P^2+Q^2}{4} \implies y^2+Qy-u-\frac{P^2}{4}=0. \tag{1} $$ The Charpit's equations corresponding to $(1)$ are $$ \frac{dx}{-\frac{P}{2}}=\frac{dy}{y}=\frac{dP}{P}=\frac{dQ}{-2y}=\frac{du}{-\frac{P^2}{2}+Qy}, \tag{2} $$ from which follow $$ P=-2(x+a)\quad\text{and}\quad Q=-2(y+b), \tag{3} $$ where $a$ and $b$ are arbitrary constants.
From $(2)$ it also follows that $du=Pdx+Qdy$; combined with $(3)$, it yields $$ u=-(x+a)^2-(y+b)^2+c. \tag{4} $$ To determine the the constant $c$, we substitute $(3)$ and $(4)$ in $(1)$; the result is $c=b^2$. Therefore, the complete integral of the original PDE is the family of spheres $$ (x+a)^2+(y+b)^2+z^2=b^2. \tag{5} $$