Lemma. Let $(d \mid \cdot ) : \Bbb{Z} \to \{0,1\}$ be whether $(1)$ or not $(0)$ $\ d$ divides the input.
There exists no $N \in \Bbb{N}$ such that $\forall x \geq N$ we have $$f(x) = \sum_{d \mid p_{\pi(\sqrt{x+1})}\#} (d \mid x^2 - 1) = 0 \pmod 2$$
This is a strange conjecture I've come up with that relates to the twin prime problem.
Attempt. Suppose the opposite, then we may augment the summation bound of $\pi(\sqrt{x + 1})$ all the way to $\pi(x-2)$, based upon a sieve-of-Eratosthenes argument. Either way, you'll get $f(x) = 0$. There are a growing number of primes that occur between $\lfloor\sqrt{x + 1}\rfloor + 1 \dots x - 2$.
If $g(x) =f(x)$ written with the next prime $q = p_{\pi(\sqrt{x + 1}) + 1}$, we have:
$$ 0 = f(x) = g(x) = (1 + (q \mid x^2-1))f(x) = f(x) + (q\mid x^2 - 1) f(x) \\ \ = 0 + \sum_{d \mid p_n\#}(qd\mid x^2 - 1), \text{ where } n \equiv \pi(\sqrt{x + 1}). $$
But we can turn this summation back from whence it came into a product:
$$ 0 = g(x) = \prod_{2 \leq p \leq p_n} (1 + (qp \mid x^2 - 1)) \pmod 2 $$
This implies that there exists $p \leq p_n$ such that $(qd \mid x^2-1)$ where $q = p_{\pi(p) + 1}$ is the next prime.
Now repeat the procedure, to get:
$$ h(x) = (1 + (r|x^2 -1))g(x) = 0 = g(x) + (r\mid x^2 - 1) g(x) \\ \ = \prod_{2 \leq p \leq p_n} ( 1 + (rqp\mid x^2 - 1) ) \pmod 2 $$
which means that there exists $p \leq p_n$ such that $rqp = p_n$ where $q, r$ respectively are the next two primes. Now do this repetitively until you use up all the primes in range $\lfloor \sqrt{x + 1} \rfloor + 1 \dots x -2$.
All together, this means:
There exists $N \in \Bbb{N}$ such that for all $x \geq N$ we have that:
$$ p\cdot \prod_{i= \pi(\sqrt{x + 1}) + 1}^{\pi(x - 2)} p_{i} $$
divides $x^2 - 1$ for some prime $p \leq \sqrt{x + 1}$. Or in terms of primorial, we have that:
$$ \exists N \in \Bbb{N}:\forall x \geq N, \exists p \leq \sqrt{x + 1}: $$
$$ p\frac{(x-2)\#}{\sqrt{x + 1}\#} \mid (x^2 -1) $$
To someone experienced with numbers, this seems like an absurdity, but I'm not quite sure how to prove it. The primorial notation is just $(x-2)\#:= p_1 p_2 \cdots p_{\pi(x-2)}$ for example.
Proving (the opposite of) this requires estimating $\pi(x-2) - \pi(\sqrt{x + 1})$ perhaps.
Or can we argue by divisibility that given $x = N$; $x^2 -1$, $(x + 1)^2 -1$ and $(x + 2)^2 - 1$ and so on can't all share the above property.
Any ideas?
First note that for any positive integer $n$ the map $$\Bbb{N}_{>0}\ \longrightarrow\ \{0,1\}:\ d\ \longmapsto\ (d\mid n),$$ is multiplicative. It follows that for any positive integer $k$ we have $$\sum_{d\mid p_k\#}(d\mid n)=\prod_{i=1}^k\left(1+(p_i\mid n)\right).$$ This shows that the sum is even if and only if $n$ is divisble by some prime $p\leq p_k$.
This shows that your conjecture is equivalent to the statement that there are infinitely many positive integers $x$ such that $x^2-1$ is not divisible by any prime smaller than $\sqrt{x+1}$. Of course $$x^2-1=(x+1)(x-1),$$ and so both $x+1$ and $x-1$ must then be prime. That is to say, your conjecture is equivalent to the twin prime conjecture.