Does the equation in positive integers $(n,y)\,\prod_{k=1}^n(p_k^2-1)=y^2\,$ only have the solution $(3,24)$?

Question

Does the equation in positive integers $\,(n,\,y)$ $$\prod_{k=1}^n(p_k^2-1)=y^2$$ only have the solution $(3,\,24)\,$?

I asked a more general question here.

The computational complexity of the problem can be slightly reduced considering that $$(2^2-1)(3^2-1)(5^2-1)=24^2$$ $$24|p^2-1\;\;\;\;\;p\gt3$$ Therefore, one can consider the equivalent problem $$\frac{\prod_{k=1}^{2n+3}(p_k^2-1)}{24^{2(n+1)}}=z^2$$ and the known solution $\,(n,\,z)=(0,\,1)$.

Up to $\,n=5\cdot10^4-2\,$ I have not found any other solution.

Many thanks.

Answer 1

Not an answer - only posting here since no pictures can be placed in comments

That the $n$th row of the above picture has black pixel at the $m$th position means that the $m$th prime has odd multiplicity in $\prod_{k\leqslant n}(\mathrm{Prime}_k^2-1)$. So a counterexample would mean having an entirely white horizontal line.

The widening black streak to the right actually comes from a string of $1$s; that is, the larger and larger amount of highest prime divisors all have multiplicity $1$. Should be not entirely impossible to show this. Actually, it obviously suffices to show that the highest prime divisor has multiplicity $1$.