Solve the equation $x^2+x+2=2^y$ in positive integers. Let $x,y$ are positive integers. Solve the equation
$$x^2+x+2=2^y$$
My work so far:
$$(2x+1)^2 + 7 = 2^{y+2}$$
If $y$ is even, then $x=1,y=2$.

Solutions are (brute force) $$(90,13);\;(5,5);\;(2,3);\;(1,2)$$
I managed to find only the even solution $y=2$ analytically 
I did in this way
Solving $x^2 + x + 2- 2^y=0$ 
discriminant is $1-4(2-2^y)=4\cdot 2^y-7$
It is a perfect square if 
$4\cdot 2^y -7=a^2$
that is $4\cdot 2^y-a^2=7$
factoring RHS
$\left(2\cdot 2^{y/2}+a\right)\left(2\cdot 2^{y/2}-a\right)=7$
Which means that $2\cdot 2^{y/2}+a=7$ and $2\cdot 2^{y/2}-a=1$
that is $a = 2\cdot 2^{y/2}-1$ and 
$2\cdot 2^{y/2}+ 2\cdot 2^{y/2}-1=7$
$ 4\cdot 2^{y/2}=8\to 2^{y/2}=2\to \color{red}{y=2;\;x=1}$
But I don't know how to find the odd solutions for $y$
Hope this helps
