Cubic diophantine equation in 3 variables $(x+2y)(x-4y+k)(x-4y-k) - 28y^3 = 0$, $x,y,z \neq 0$ From research completely unrelated to Number Theory I stumbled onto the following equation:

$$
xyz = \frac{7}{16}\left(\frac{2x - y - z}{3}\right)^3
$$

for $x, y, z$ integers, $x,y,z \neq 0$(I forgot to put this condition in the original question). Are there integers that satisfy it?

Edit: Ok, I was missing the condition that all variables should be nonvanishing. Here is the original problem

$$
(x+2y)(x-4y+z)(x-4y-z) - 28y^3 = 0
$$

for $x,y,z$ integers and (all of them) nonvanishing. 

The first equation can be obtained from this one by a trivial change of variables. Of course, solutions in the integers for the first one do not imply solutions in the integers for the second one but that is fine because rational solutions are also OK.
By expansion and simplification, your original problem can be written as
$$
  2(3x^2-2y^2+z^2)y=x(x-z)(x+z).
$$
Let $d = \gcd(x,y)$, say $(x,y)=(du,dv)$ for integers $(u,v)$ with $\gcd(u,v)=1$. Then
$$
  2\bigl((3u^2-2v^2)d^2+z^2\bigr)v=u(du-z)(du+z).
$$
Now re-simplifying in terms of $d$,
$$
d^2(u^3-6u^2v+4v^3) =(2v+u)z^2.
$$
Can you take it from there?
There are lots of easy solutions where one of $x,y,z$ is $0$.