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?

$x^3+x^2=2^y+16$. 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 $(x,y)=(4,6)$.
