I believe the general solution to $x^2+y^2-z^2=1$ is $x=(rs+tu)/2$, $y=(rt-su)/2$, $z=(rs-tu)/2$, where $rt+su=2$.
EDIT: Solutions to $x^2+y^2+1=z^2$ can be obtained by choosing $a$, $b$, $c$, $d$ such that $ad-bc=1$ and then letting $x=(a^2+b^2-c^2-d^2)/2$, $y=ac+bd$, $z=(a^2+b^2+c^2+d^2)/2$, though I'm not sure you get all the integer solutions this way.
The rational solutions are a bit easier. $(0,0,1)$ is a (rational) point on the surface. The line $(0,0,1)+t(a,b,c)$ through that point intersects the surface again at $x=2ac/(a^2+b^2-c^2)$, $y=2bc/(a^2+b^2-c^2)$, $z=(a^2+b^2+c^2)/(a^2+b^2-c^2)$, giving all the rational points on the surface.