Using a solution to show no others are possible in positive odd integer

Question

If I have an equation say $$3(1+x+x^2)(1+y+y^2)(1+z+z^2)+1=4x^2y^2z^2 \quad (1)$$ and I know a non negative integer solution $x=4, y=64, z=262144$ then no odd positive integer solutions can possibly exist. I know that one way to show it for this example would be to compute but I am looking for a proof that does not rely on computation like using wolfram as I want to generalize this. I will happily award the bounty to anyone who can prove it. What I have tried is minimal. I suppose that $p,q,r \in \mathbb N$ and they satisfy equation $(1)$. I think if we let $p$ be the smallest integer of the solution $p<4$ is impossible so assume that $p\ge 5$ but this might lead to a contradiction since the coefficient of $4p^2q^2r^2 \quad $ is $4<5$ and that might be impossible?

Answer 1

The given equation is $3(1+x+x^2)(1+y+y^2)(1+z+z^2)=4x^2y^2z^2-1$.

Let $x \leq y \leq z$. Then $x=1$ has no solution as in this case the LHS is at least $9y^2z^2>4y^2z^2$. Thus, let $x \geq 3$. Also note that $3$ divides the LHS so $3$ does not divide $xyz$; hence $x \geq 5$.

Divide the original equation by $x^2y^2z^2$ on both sides and upper-bound each sum by an infinite geometric series to get:

$3\dfrac{xyz}{(x-1)(y-1)(z-1)}>4-\dfrac{1}{x^2y^2z^2}>4-\dfrac{1}{xyz(x-1)(y-1)(z-1)}$.

Thus, $3xyz\geq 4(x-1)(y-1)(z-1)$, so that $$\left(1-\dfrac{1}{x}\right)\left(1-\dfrac{1}{y}\right)\left(1-\dfrac{1}{z}\right)\leq\dfrac{3}{4}.$$

From the above, the minimum of $x,y,z$ must be at most $9$, otherwise the LHS is at least $(10/11)^3>3/4$. This means that $x \in \{5,7\}$.

If $x=5$, then $y \leq 29$ and if $x=7$, then $y \leq 15$, both upper bounds obtained from the previous inequality.

When $x=5$, considering the cases modulo 5, we must have $y \equiv 0/2/4$ (mod 5). Thus, we are finally left with the following pairs of $(x,y)$ to check and eliminate:

$(5,5),(5,7),(5,17),(5,19),(5,25),(5,29),(7,7),(7,13),(7,15)$.