Suppose that it is not so. Consider the sequence $(x_n)$ of all members, the indices of which do not contain in their decomposition on factors the squares of prime numbers. Then in this subsequence there is only a finite number of prime divisors $p_1, p_2, \dots, p_k$ other than $2$, $3$, $7$ (prime factors of the number $2016$). Let $n_0 = \prod_{i=1}^{k} p_i^{\alpha_i}$, where $\alpha_i \ge x_1$. Consider the following sequence:

$a_n = 1 + n n_0 \phi(n_0)$, where $\phi(n)$ is Euler function, i.e. the function that counts the positive integers up to a given integer $n$ that are relatively prime to $n$. Euler function can be represented in the form of the so-called Euler product: $\phi(n) = n \prod_{p|n} (1 - \frac{1}{p})$, where $p$ is a prime number.

It is clear that $Q(a_n) \equiv Q(1) \pmod{n_0}$ and $P(a_n) \equiv P(1) \pmod{\phi(n_0)}$. As by the construction $(n_0, 2016) = 1$, hence $x_{a_n} = 2016^{P(a_n)} + Q(a_n) \equiv x_1 \pmod{n_0}$.

Later we will show that for large $n \ge 1$ and for each $p_i$, $i = 1, k$ the sequence of values $\text{ord}_{p_i}(x_{a_n})$ is limited. We can choose an infinite number of primes from the sequence $(a_n)$, which follows from Dirichlet's theorem or from using simple arguments about the progression with the first member $1$. Next, let us consider only those $n$ for which corresponding $a_n$ is a prime number. Then we have that there are infinitely many values $n \ge 1$:

$$2016^{P(n)} + Q(n) = 2^a \cdot 3^b \cdot 7^c \cdot M, \quad (1)$$
for a fixed limited $M$. But then
$$M \cdot (\max\{2^a; 3^b; 7^c\})^3 \ge M \cdot 2^a \cdot 3^b \cdot 7^c = 2016^{P(n)} + Q(n) > 2016^n.$$
Then from one of these prime numbers $A$ (which is the largest in the previous transition) in equation (1) we have that is $2016^{P(n)}$ divisible by $A^{\frac{1}{3}n-C_0}$. The right-hand side is also divisible by $A^{\frac{1}{3}n-C_0}$, hence $Q(n)$ should be divisible by the same power. However, for sufficiently large values $n$ it is not possible, because $Q(n)$ is a power function, while $A^{\frac{1}{3}n-C_0}$ is exponential.