= Solution
Put $W=1+x+y$. The identity $W=(s+1)^2$ gives $s=-1\pm\sqrt W$, and hence $z=-1\pm2\sqrt W$. On the initial half-line $W=1$ and the prescribed value is $z=1$, which selects the positive sign. The branch containing the data has $s>-1$, so
$$
\boxed{z(x,y)=-1+2\sqrt{1+x+y}.}
$$
For $W>0$, its <derivatives> are $z_x=z_y=W^{-1/2}$. Substitution checks the <partial differential equation>:
$$
z_x+zz_y=\frac{1+z}{\sqrt W}=2.
$$
On the data $y=-x$, the value is one. The other algebraic branch gives $z=-3$ there and does not meet these data. The positive-branch formula defines a solution on the larger half-plane $W>0$, but the part uniquely fixed by the prescribed half-line also requires a nonnegative characteristic footpoint $t$, as part (vi) explains.
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