$x = hz; y = kz$ h,k are coprimes. $h + zk^2 = z^2(hk - 1)$ $z | h+zk^2 \ ; \ z| \frac hz + k^2 $ $z | h \ ; \ h=gz$ Then, $z | g + k^2$ Since h and k were coprimes, $z !| k$ and $z !| k^2$ Therefore $g = fz$ Original Equation becomes $$z(fkz^2-f-1) = k^2$$h and k were coprimes and $z|h$, Therefore $z !| k$ Only possibility if $$z=1 \ ; \ fk - f - k^2 - 1 = 0$$, Taking quadratic in k, $k = \frac {f \pm \sqrt{(f-2)^2-8}}2$ $(f-2)^2-8 = a^2$ $$8 = (f-2-a)(f-2+a) ; (f,k,z) = (f, \frac {f \pm a}2,1)$$$(a+f-2,f-2-a) = [(2,4),(4,2),(-2,-4),(-4,-2)]$, we take only same parity duples. $(f,a)=[(5,-1),(5,1),(-1,1),(-1,-1)]$ And $f=a \mod 2$ for k to be an integer and $f>0$ $(f,k)=[(5,1),(5,-1)]$ and k>0 given. And also f and k have to be coprimes as h and k are coprimes and z=1, So $$(f,k,z) = [(5,3,1),(5,2,1)]$$ $$(x,y,z) = [(5,3,1),(5,2,1)]$$are the $Z^+$ solutions to $x + y^2 + z^3 = xyz$ where $z = gcd(z,y)$. [edit: Ive missed the Even Cases Somehow]