Integer solutions to multivariate polynomial I'm looking for a way to show that an equation like  $$15(1+x+x^{2})(1+y+y^{2})=16(xy)^{2}-1$$ has no solutions with x and y odd positive integers. The 16 and 15 can be changed  but I'm trying to generalize it. Any help is appreciated.

Factorization should be useful, but another idea is to note that
if the equation is $a(x^2+x+1)(y^2+y+1)=b(xy)^2-1$ and $a<b$, then $min(x,y)$ is upper-bounded by a function of $a,b$ and it is sufficient to examine those values (and solve for the other).
Specifically, here is one (not necessarily sharp) bound: $min(x,y) \leq \dfrac{\sqrt{b}}{\sqrt{b}-\sqrt{a}}$.
To prove the bound, we write: $a(1+\dfrac{1}{x}+\dfrac{1}{x^2})(1+\dfrac{1}{y}+\dfrac{1}{y^2})=b-\dfrac{1}{x^2y^2}$;
thus $a\dfrac{xy}{(x-1)(y-1)}>b-\dfrac{1}{xy(x-1)(y-1)}$,
and hence $axy>b(x-1)(y-1)-1$, so that $\left(1-\dfrac{1}{x}\right)\left(1-\dfrac{1}{y}\right)\leq\dfrac{a}{b}$.
So if $x,y \geq n$, then $1-1/n \leq \dfrac{\sqrt{a}}{\sqrt{b}}$ and thus $n \leq \dfrac{\sqrt{b}}{\sqrt{b}-\sqrt{a}}$.
