diophantine question parity and sign If I have an equation say $$3(1+x+x^2)(1+y+y^2)=4x^2y^2-1 \quad (1)$$ and I rewrite it as 
$$4+3(x+y+xy+x^2+y^2+xy^2+x^2y)-x^2y^2=0$$ and I find a positive integer solution where both x and y are even, can I show that no odd positive integer solutions exist?   

Since the equation is symmetric in $x$ and $y$, we only need to consider the case $x\le y$.
Dividing the both sides of
$$3(1+x+x^2)(1+y+y^2)=4x^2y^2-1\tag1$$
by $x^2y^2$ gives
$$3\bigg(1+\frac 1x+\frac{1}{x^2}\bigg)\bigg(1+\frac 1y+\frac{1}{y^2}\bigg)=4-\frac{1}{x^2y^2}\tag2$$
Suppose here that $x\ge 8$.
Then, we have
$$ \text{(LHS of $(2)$)}\le 3\bigg(1+\frac 18+\frac{1}{8^2}\bigg)\bigg(1+\frac 18+\frac{1}{8^2}\bigg)=\frac{15987}{4096}$$
and
$$\text{(RHS of $(2)$)}\ge 4-\frac{1}{8^4}=\frac{16383}{4096}$$
This is a contradiction. So, we have $x\le 7$.
Now, $(1)$ can be written as
$$(3+3x-x^2)y^2+3(1+x+x^2)y+4+3x+3x^2=0$$
Seeing this as a quadratic equation on $y$ and considering the discriminant, we get
$$D=9(1+x+x^2)^2-4(3+3x-x^2)(4+3x+3x^2)$$
It is necessary that $D$ is a square number.
For $x=1$, we get $D=-119$ which is not a square number.
For $x=3$, we get $D=1041$ which is not a square number.
For $x=5$, we get $D=11281$ which is not a square number.
For $x=7$, we get $D=46441$ which is not a square number.
Therefore, $(1)$ has no solution $(x,y)$ such that both $x$ and $y$ are positive odd integers.

Added : 
For $x=2$, $D$ is a square number, but $y=-2$ is not positive.
For $x=4$, $D$ is a square number, and $y=-1,64$.
For $x=6$, $D$ is not a square number.
Therefore, $$\color{red}{(x,y)=(4,64),(64,4)}$$ are the only positive integer solutions.
