Solution:

The key fact is that, if $a+1=u^{2}$ and $b=v^{2}$ are perfect squares and $a>b$, then $a b+1$ is not a square. Indeed, then $(u v-1)^{2}<a b+1=u^{2} v^{2}-v^{2}+1<(u v)^{2}$.

Let $p_{1}, p_{2}, \ldots, p_{m}$ be all the primes less than 2016. For $1 \leqslant n \leqslant 2^{2016}$ consider the binary sequence $\mathrm{c}_{n}=\left(r_{1}, r_{2}, \ldots, r_{m}\right)$, where $r_{i}=0$ if the exponent of $p_{i}$ in the product $P_{n}=a_{1} a_{2} \cdots a_{n}$ is even, and $r_{i}=1$ otherwise. Since there are only $2^{m}$ possibilities for the sequence $c_{n}$, for every $k \leqslant 2^{2016}-2^{m}$ among the sequences $c_{k+1}, \ldots, \mathrm{c}_{k+2^{m}}$ there exist two that are equal, say $c_{s}$ and $c_{t}$ ($s<t$), and then $P_{t} / P_{s}$ is a perfect square not greater than $2016^{2^{m}}$.

Suppose that in the sequence $\left(a_{n}\right)$ there are no ones. Take $k=11 \cdot 2^{m}$; certainly $k+2^{m}<2^{2016}$. We have seen that there exist indices $s, t, k \leqslant s<t \leqslant k+2^{m}$, such that $b=P_{t} / P_{s}$ is a perfect square. However, since $a=P_{s} \geqslant P_{k} \geqslant 2^{k}=2048^{2^{m}}>2016^{2^{m}} \geqslant b$, by the fact from the beginning, $a+1=P_{s}+1$ and $a b+1=P_{t}+1$ cannot simultaneously be squares, which is a contradiction.