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).

etc.

Obviously, none of these are among the 4 smallest