I noticed something as I was playing around with prime numbers. By denoting $p_i$ the $i^{\text{th}}$ prime number, I discovered the following:
$$ \begin{align}\prod_{i=1}^2\left(p_i^{ \ 2}+i\right)&= 28^2-27^2 \\ \prod_{i=1}^5\left(p_i^{ \ 2}+i\right)&=3207^2-27^2.\end{align} $$
Given the general equation
$$\prod_{i=1}^m\left(p_i^{ \ 2}+i\right)=n^2-27^2\tag*{$\big(\left(m,n\right)\in\mathbb{N}^2\big)$}$$
I have tested for $5< m\leqslant 32$ and it appears that for $m > 5$, there does not exist such $n$. I know that $n^2-27^2=(n+27)(n-27)$, but this does not help me make much progress.
Is this true? Can it be verified? Thus far, it remains a conjecture that the only solutions $(m,n)$ are $(2,28)$ and $(5,3207)$.
Source of Testing:
$$\text{Alpertron $-$ Integer Factorization Calculator.}$$
Thank you in advance.