Find all triples of positive integers $(x, y, z)$ satisfying  
$$ 2(x+y+z+2 x y z)^{2}=(2 x y+2 y z+2 z x+1)^{2}+2023 $$
Answer: $(3,3,2)$ and permutations. Claim - The identity 
$$ 2(x+y+z+2 x y z)^{2}-(2 x y+2 y z+2 z x+1)^{2}=\left(2 x^{2}-1\right)\left(2 y^{2}-1\right)\left(2 z^{2}-1\right) $$
is true.
$$ \operatorname{Norm}(a+b \sqrt{2})=a^{2}-2 b^{2} $$
which is multiplicative, meaning
$$ \operatorname{Norm}(u \cdot v)=\operatorname{Norm}(u) \cdot \operatorname{Norm}(v) $$
This means that for any rational numbers $x, y, z$, we should have
$$ \begin{aligned} & \operatorname{Norm}((1+\sqrt{2} x)(1+\sqrt{2} y)(1+\sqrt{2} z)) \\ = & \operatorname{Norm}(1+\sqrt{2} x) \cdot \operatorname{Norm}(1+\sqrt{2} y) \cdot \operatorname{Norm}(1+\sqrt{2} z) \end{aligned} $$
But $(1+\sqrt{2} x)(1+\sqrt{2} y)(1+\sqrt{2} z)=(2 x y+2 y z+2 z x+1)+(x+y+z+2 x y z) \sqrt{2}$ so the above equation is the negative of the desired identity.
We are thus reduced to find positive integers $x, y, z$ satisfying
$$ \left(2 x^{2}-1\right)\left(2 y^{2}-1\right)\left(2 z^{2}-1\right)=2023=7 \cdot 17^{2} $$
Each of the factors is a positive integer greater than 1. The only divisors of 2023 of the form $2 t^{2}-1$ are $1,7,17$. This gives the answers claimed.