diophantine equation $x^3+x^2-16=2^y$ Solve in integers: $x^3+x^2-16=2^y$.
my attempt:
of course $y\ge 0$, then $2^y\ge 1$, so $x\ge 1$.
for $y=0,1,2,3$ there is no good $x$.
so $y\ge 4$ and we have equation $x^2(x+1)=16(2^z+1)$, where $z=y-4\ge 0$.
what now?
 A: This is only a solution for even values of $x$:
For $x$ even, $x^2$ is even and $x+1$ is odd. So $\gcd(16, x+1) = 1$. So $16 \mid x^2$. So $4 \mid x$. So $x = 4k$ for some $k$.
$16k^2(4k+1)=16(2^z+1)$
$k^2(4k+1)=2^z+1$
If $k$ is even then the RHS is even and we have a contradiction. So $k$ must be odd. So $k = 2t+1$ for some $t$.
$(2t+1)^2 (8t+5) = 2^z + 1$
$(4t^2 + 4t + 1)(8t + 5) = 2^z + 1$
$32t^3 + 52t^2 + 28t + 4 = 2^z$
We see $z \geq 2$. Let $w = z - 2$.
$8t^3 + 13t^2 + 7t + 1 = 2^w$
For $w \geq 1$, $t^2 + t + 1 \equiv 0 \pmod{2}$ which has no solution in $t$.
So $w = 0$.
$8t^3 + 13t^2 + 7t = 0$
$t(8t^2 + 13t + 7) = 0$
Take the discriminant of the quadratic, so we get $169 - 224 < 0$.
So $t = 0$ and $w = 0$.
So for $x$ even, $x = 4k = 4(2t + 1) = 4$ and $y = z + 4 = w + 6 = 6$.

You can also bound (bind?) $x$, giving it in terms of $y$. Go back to $x^3+x^2−16=2^y$. Consider $f(x,y) = x^3 + x^2 - 16 - 2^y$. Evaluate $f(2^{y/3},y) = 2^y + 2^{2y/3} - 16 - 2^y = 2^{2y/3} - 16$ which equals $0$ when $y = 6$ and is bigger than $0$ when $y > 6$. Evaluate $f(2^{y/3} - 1, y)$ which you can check here is less than $0$ for all $y$. This shows that if $x$ is an integer and $f(x,y) = 0$ then $x = \left \lfloor 2^{y/3} \right \rfloor$.
As far as working with $x = \left \lfloor 2^{y/3} \right \rfloor$, you can write it as $x = \left \lfloor 2^q2^{r/3} \right \rfloor$ where $0 \leq r < 3$  which can be expressed as $\left \lfloor 10_2^q 10_2^{r/3} \right \rfloor$ in binary. As $q$ increases (which increases $y$ by 3), the value of $x$ either gets doubled, or is double plus one. I'm not sure if that's useful.
How to finish this?
A: Not my answer. Copied.
$x^2(x+1) = 2^y + 16$
Since LHS is an integer, then we must have $y \ge 0$.
Since RHS is a positive integer, then we must have $x \ge 1$.
$x^2(x+1)$ is strictly increasing for $x \ge 0$.
$2^y + 16$ is strictly increasing for $y \ge 0$.
$$\begin{array}{n|c|c|}
\hline
 n & n^2(n+1) & 2^n + 16 \\ \hline
0 & 0 & 17 \\
1 & 2 & 18 \\
2 & 12 & 20 \\
3 & 36 & 24 \\
4 & 80 & 32 \\
5 & 150 & 48 \\
6 & 252 & 80\\ \hline
\end{array}$$
Note that the table indicates that $(x,y) = (4,6)$ is a solution.
So any solution involving $y \ge 7$ will require $x \ge 5$.
We will show that there is no solution for $y \ge 7$.
So we can assume now  that $x \ge 5$ and $y \ge 7$.
\begin{align}
    x^2(x+1) &= 2^y + 16\\
    x^2(x+1) &= 16(2^{y-4} + 1)\\
\end{align}
Note that $2^{y-4}+1$ must be an odd integer.
So if $x$ is an odd integer, $\gcd(x+1,2^{y-1}+1) = 1$.
$\quad$ Hence $x+1 | 16$
$\quad$ Remembering that $x \ge 5$, we must have $x = 7$ or $x = 15$.
$\quad$Case $1: x = 7$
\begin{align}
  16(2^{y-4}+1) &= 392\\
  2(2^{y-4}+1) &= 49 & \text{Has no solution.}\\
\end{align}
$\quad$Case $2: x = 15$
\begin{align}
  16(2^{y-4}+1) &= 3600\\
  2^{y-4}+1 &= 225 & \\
  2^{y-4} &= 224 \\
  2^{y-4} &= 32 \times 7 & \text{Has no solution.}\\
\end{align}
So if $x$ is an even integer, $\gcd(x^2,2^{y-1}+1) = 1$.
$\quad$ So $x^2 | 16$. 
This can't happen since $x \ge 5$.
A: $x^3+x^2=2^y+16$. The RHS is positive, so $x^2(x+1)>0\iff x\ge 1$. Since $2^y$ is an integer, we have $y$ is a positive integer too ($y=0$ won't give a solution).  
$x,y$ are positive integers.
$x^3+x^2-16=2^y$. You see a cubic polynomial on the LHS that could easily be strictly bounded between two consecutive cubes (namely $x^3$ and $(x+1)^3$) for most values of $x$, making it impossible for it to be a cube itself. So if $2^y$ is a cube, i.e. $3\mid y$, we're done. And indeed it is a cube.  
$2^y\equiv 1,2,4\pmod{7}$ for $y\equiv0,1,2\pmod 3$, respectively.   
$x^3+x^2-16\equiv 5,0,3,6,1,1,5\pmod{7}$ for $x\equiv 0,1,2,3,4,5,6\pmod{7}$, respectively.  
The only common residue is $1$, so $y\equiv 0\pmod{3}$. This implies $x^3+x^2-16$ is a cube.  
But $x^3<x^3+x^2-16<(x+1)^3,\forall x\ge 5$, so $1\le x\le 4$, which only give $(x,y)=(4,6)$.
