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!
 A: Render the equation as
$x^2+7x+(4-2^y)=0$
and treat this as a quadratic equation for $x$. If this is to have natural number roots the discriminant must be a perfect square, which you can work out:
$\Delta=33+(4×2^y)=m^2$
where wlog $m$ may be rendered positive.
If $y$ is odd, then $2^y\equiv2\bmod3$, and substituting that into the discriminant above then gives $m^2\equiv2\bmod3$. That clearly fails, so $y$ must be even and we put in $y=2z$.
Then we have
$33+(4×2^y)=m^2$
$4×2^y=(2×2^z)^2=n^2, n=2×2^z$
where we also render $n$ positive.
Thus by difference
$m^2-n^2=(m+n)(m-n)=33,$
and evidently we must take one of the following:
$m+n=33,m-n=1\implies m=17,n=16$
$m+n=11,m-n=3\implies m=7,n=4$
Comparing the values of $n$ with $n=2×2^z$ then gives $z\in\{1,3\}$, from which $y=2z\in\{2,6\}$ and the complete solution set follows.
A: If we take mod 3 and $y=2z+1$ then $2^y\equiv -1$ but $$x^2+7x+4 \equiv  0,1$$ a contradiction, so $y=2z$. Let $2^z = t$, then we have

*

*If $x>0$ then

$$(x+2)^2\leq \underbrace{x^2+7x+4}_{t^2} \leq (x+4)^2 \implies t = x+3\implies x=5$$

*

*If $x<0$ then $x=-n$ with positive $n$. So $$(n-4)^2\leq \underbrace{n^2-7n+4}_{t^2} \leq (n-2)^2 \implies t = n-3\implies n=-5$$so no solution.


*If $x=0$ then $y= 2$.
A: We know that: $x=\frac{-7\pm\sqrt D}2$
For natural solutions: $$x=\frac{-7+\sqrt D}2$$
Also, $D=n^2$ where $n$ is an natural number. And, $n$ has to be odd and $n\geq7$.
$$\Rightarrow 7^2-4(4-2^y)=n^2$$
$$\Rightarrow 7^2-16-2^{y+2}=n^2$$
$$\Rightarrow n^2-33=2^{y+2}$$
We put $n=7$ (our first choice), and get $y=2$ (and $x=\frac{n-7}2$).
Now we don't have to check for $n=9,11$ or $13$ as for $y\geq2, \ (n^2-33)$ must be divisible by $16$, so, $n=16k\pm1$.
Check for $n=15:2^{y+2}=192\Rightarrow y\notin N.$ 
Check for $n=17:2^{y+2}=256\Rightarrow y=6.$ 
Now we know that our next solution will be for $y>6\Rightarrow2^{y+2}>256$ which means that $n$ divided by $256$ should leave a remainder of $\sqrt33$ which is not possible.
So,  the required answer: $(x,y)=(\frac{7-7}2,2)$ and $(\frac{17-7}2,6)$.
$$\bf \text{Thus, }\boxed{(x,y)=(0,2),(5,6)}$$.
A: 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.
