Problem: Find all triples of positive integers $(x,y,z)$ satisfying
\[ 2(x+y+z+2xyz)^2 = (2xy+2yz+2zx+1)^2 + 2023. \]

Solution: Answer: $(3,3,2)$ and permutations.

The solution hinges upon the following claim:
\begin{claim*}
   The identity
   \[ 2(x+y+z+2xyz)^2 - (2xy+2yz+2zx+1)^2 = (2x^2-1)(2y^2-1)(2z^2-1) \]
   is true.
\end{claim*}
\begin{proof}
  This can be proved by manually expanding; we show where it ``came from''.
  In algebraic number theory, there is a \emph{norm function}
  $\opname{Norm} \colon \QQ(\sqrt2) \to \QQ$
  defined by
  \[ \opname{Norm}(a+b\sqrt2) = a^2-2b^2 \]
  which is \emph{multiplicative}, meaning
  \[ \opname{Norm}(u \cdot v) = \opname{Norm}(u) \cdot \opname{Norm}(v). \]

  This means that for any rational numbers $x$, $y$, $z$, we should have
  \begin{align*}
    &\phantom{=} \opname{Norm}
    \left( (1+\sqrt2x)(1+\sqrt2y)(1+\sqrt2z) \right) \\
    &= \opname{Norm}(1+\sqrt2x)
      \cdot \opname{Norm}(1+\sqrt2y)
      \cdot \opname{Norm}(1+\sqrt2z).
  \end{align*}
  But $(1+\sqrt2x)(1+\sqrt2y)(1+\sqrt2z) = (2xy+2yz+2zx+1) + (x+y+z+2xyz)\sqrt2$
  so the above equation is the negative of the desired identity.
\end{proof}

We are thus reduced to find positive integers $x$, $y$, $z$ satisfying
\[ (2x^2-1)(2y^2-1)(2z^2-1) = 2023 = 7 \cdot 17^2. \]
Each of the factors is a positive integer greater than $1$.
The only divisors of $2023$ of the form $2t^2-1$ are $1$, $7$, $17$.
This gives the answers claimed.