Determine all pairs $(x,y)$ of integers

Question

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

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

Answer 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}