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

I tried:
$$1+2^x+2^{2x+1}=y^2$$
$$2^x+2^{2x+1}=y^2-1$$
$$2^x+2^x2^x2=(y+1)(y-1)$$
$$2^x(1+ 2^{x+1})=(y+1)(y-1).$$

Completing the square gives
\begin{equation}
\left(2^{x+2}+1\right)^2=8y^2-7
\end{equation}
so $8y^2-7$ must be a perfect square of a number which is one more than four times a power of two. So two solutions are $(0,2)$ and $(0,-2)$.
Two more solutions are $(4,\pm23)$.
The steps to completing the square are as follows:
\begin{eqnarray}
2\cdot2^{2x}+2^x&=&y^2-1\\
16\cdot2^{2x}+8\cdot2^x&=&8y^2-8\\
16\cdot2^{2x}+8\cdot2^x+1&=&8y^2-7\\
\left(4\cdot2^x+1\right)^2&=&8y^2-7\\
\left(2^{x+2}+1\right)^2&=&8y^2-7
\end{eqnarray}
