Find natural number $x,y$ satisfy $x^2+7x+4=2^y$ 
Find natural number $x,y$ satisfy $x^2+7x+4=2^y$

My try: I think $(x;y)=(0;2)$ is only solution. So I try prove $y\ge3$ has no solution, by $(x+1)(x+6)-2=2^y$.
So $2\mid (x+1)(x+6)$, but this is wrong. Done.
This is wrong
Anyone has an idea? Please help, thank you!

HINT:
We have
$$x^2+7x+(4-2^y)=0\\
\Delta_x=2^{y+2}+33=z^2\\
$$
Let, $y=2k$, then
$$\left(z-2^{k+1}\right)\left(z+2^{k+1}\right)=33$$
and so
$$
\begin{cases}z- 2^{k+1}=\pm 1,3,11,33\\
z+ 2^{k+1}=\pm 33,11,3,1\end{cases}
$$
Then, let $y=2k-1$ we have
$$2^{2k+1}+1=(z^2+1)-33$$
Since, $2^{2k+1}+1 \mod 3=0$, putting $z=3n\pm 1 $ and $z=3n$ we get $z^2+1\mod 3 \neq 0$. Thus, this case is impossible.
