<s>[INST] 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!
 [/INST] 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.
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