Find a positive integer solution to $xyzw=504(x^2+y^2+z^2+w^2)$ 
Find positive integer values of $x,y,z,w$, such that $$xyzw=504(x^2+y^2+z^2+w^2)$$

I found it at some point and now I am unable to find the solution anymore, maybe this equation isn't satisfiable? But how do one prove such thing?
Edit @Dashisan found nice example $(x,y,z,w)=(21,63,84,84)$. Now, the problem is if there  exists distinct positive integer solution.
 A: Using a Pell-like equation, there are in fact infinitely many positive integer solutions to,
$$xyzw=504(x^2+y^2+z^2+w^2)\tag1$$

$1$st family:

$$\big(x,y,z,w\big)=\big(84,\;84,\;21q,\;21(4p+7q)\big)$$
where, $$p^2-3q^2=-2\tag2$$ 
An initial point is $(p,q) = (-1,1)$ yielding the OP's known $\big(x,y,z,w\big)=\big(84,\;84,\;21,\;63\big)$. As $(2)$ has infinitely many positive integer solutions, then so does $(1)$.

$2$nd family:

$$\big(x,y,z,w\big)=\big(36,\;36,\;9(4p+5q),\;9(8p+11q)\big)$$
where, $$p^2-2q^2=14\tag3$$ 
An initial point is $(p,q) = (4,-1)$ yielding $\big(x,y,z,w\big)=\big(36,\;36,\;99,\;189\big)$. And infinitely more.
(Added a day later.)

More generally:

$$\big(x,y,z,w\big)=\big(12c,\;12c,\;c(ap+7bq-ac^2q),\;c(bp-7aq+bc^2q)\big)$$
with the Pell-like,
$$\beta\, p^2-\beta(c^4-49)\,q^2=7\times 288=\color{blue}{2016}$$
where $\beta=-7 a^2 - 7 b^2 + 2 a b c^2$ for arbitrary $a,b,c$ and we recover the $2016$ asked by the OP. The first two families were just special cases.
A: $(x, y, z, w) = (21, 63, 84, 84)$ works.
A: Tell you what. It happens that
$$ 2xyzw = x^2 + y^2 + z^2 + w^2 $$
is not possible in positive integers. The way to approach this successfully is what they call Vieta Jumping. 
See what you can do. 
