Problem: Determine all pairs $(x,y)$ of integers such that
\[ 1 + 2^x + 2^{2x+1} = y^2. \]

Solution: Answers: $(0, \pm 2)$, $(4, \pm 23)$, which work.

Assume $x \ge 4$.
\[ 2^x \left( 1 + 2^{x+1} \right)
  = 2^x + 2^{2x+1} = y^2 - 1 = (y-1)(y+1). \]
So either:
\begin{itemize}
  \ii $y = 2^{x-1} m + 1$ for some odd $m$, so
  \[ 1 + 2^{x+1} = m\left( 2^{x-2}m+1 \right)
    \implies 2^x = \frac{4(1-m)}{m^2-8}. \]
  \ii $y = 2^{x-1} m - 1$ for some odd $m$, so
  \[ 1 + 2^{x+1} = m\left( 2^{x-2}m-1 \right)
    \implies 2^x = \frac{4(1+m)}{m^2-8}. \]
\end{itemize}
In particular we need $4|1 \pm m| \ge 2^4 |m^2-8|$,
which is enough to imply $m < 5$.
From here easily recover $x = 4$, $m = 3$ as the last solution
(in the second case).