I was trying to find all triples of integers $(x,y,z)$ which satisfy the equations $$x^2-y-2z=4,$$ $$y^2-2z-3x=-2,$$ $$2z^2-3x-5y=-22.$$ I got through trial and error that one triple should be $(3,3,1)$. But I can neither find other integer solutions, nor prove that it's the only one.
Any idea is appreciated.
An elegant solution is indicated in the comments.
Still, we can just find out all complex solutions. The first equation gives $y=x^2-2z-4$, so that we obtain two quadratic equations in $x$ and $z$. Taking the resultant of these two polynomials, we obtain $$ (x-3)(x^6 + 6x^5 + 11x^4 - 6x^3 + 23x^2 + 192x + 348)=0. $$ The polynomial of degree $6$ has no real root. In particular, $(x,y,z)=(3,3,1)$ is the only real solution.