<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] 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 $(\frac{17-7}2,6)$.
$$\bf \text{Thus, }\boxed{(x,y)=(0,2),(5,6)}$$.
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