Let $a$, $b$, $c$ be integers and $n$ be a positive integer such that
$$(a-b)(b-c)(c-a) + 4 = 2 \cdot 2016^n.$$
We set $a-b = -x$, $b-c = -y$ and we rewrite the equation as
$$xy(x+y) + 4 = 2 \cdot 2016^n. \tag{1}$$
If $n > 0$, then the right hand side is divisible by $7$, so we have that
$$xy(x+y) + 4 \equiv 0 \pmod{7} \tag{2}$$
or
$$3xy(x+y) \equiv 2 \pmod{7} \tag{3}$$
or
$$(x+y)^3 - x^3 - y^3 \equiv 2 \pmod{7}. \tag{4}$$
Note that, by Fermat's Little Theorem, for any integer $k$ the cubic residues are $k^3 \equiv -1, 0, 1 \pmod{7}$.
It follows that in (1) some of $(x+y)^3$, $x^3$ and $y^3$ should be divisible by $7$.
But in this case, $xy(x+y)$ is divisible by $7$ and this is a contradiction.
So, the only possibility is to have $n = 0$ and consequently, $xy(x+y) + 4 = 2$, or, equivalently,
$$xy(x+y) + 4 = -2. \tag{7}$$
The solutions for this are $(x, y) \in \{(-1, -1), (2, -1), (-1, 2)\}$,
so the required triples are $(a, b, c) = (k+2, k+1, k)$, $k \in \mathbb{Z}$, and all their cyclic permutations.

Alternative version:
If $n > 0$ then $9$ divides $(a-b)(b-c)(c-a) + 4$, that is, the equation
$$xy(x+y) + 4 \equiv 0 \pmod{9}$$
has the solution $x = b-a$, $y = c-b$.
But then $x$ and $y$ have to be $1$ modulo $3$, implying $xy(x+y) \equiv 2 \pmod{9}$, which is a contradiction.
We can continue now as in the first version.