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.
The solution hinges upon the following claim:
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.
Proof. This can be proved by manually expanding; we show where it "came from". In algebraic number theory, there is a norm function Norm: $\mathbb{Q}(\sqrt{2}) \rightarrow \mathbb{Q}$ defined by
$$
\operatorname{Norm}(a+b \sqrt{2})=a^{2}-2 b^{2}
$$
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.