Actually, there are an infinite number of solutions in positive integers.
Starting with Jer's y=x(+/-)x*sqrt(2016/x)
The (-) gives positive solutions if x = 2016*n^2
Then y = 2016(n^2 - n)
z = 2016(n^2 - 2n)
Checking, we see that 2016*(n^2, n*(n-1), (n-1)^2) is indeed a geometric progression, with ratio (n-1)/n
In order for z to be positive, n must be >= 3
The first of these solutions is 2016*(9,6,3).
The next is 2016*(16,12,8).
Obviously, none of these are among the 4 smallest