I was looking at perfect numbers and came across something that might serve a little interesting.
Denote by $P_n$ the $n^\text{th}$ perfect number, then there appears to always exist $k\in\mathbb{W}$ and $l\in\mathbb{Z}$ such that
$$P_n^{\,2}+5^2+2^k=(P_n-1)^2+l^2.\tag{$\verb|Conjecture|$}$$
for which $\mathbb{W}:= \mathbb{N}\cup\big\{0\big\}=\big\{0,1,2,3,\ldots\bigr\}$; id est, the whole numbers.
Here is what I have discovered thus far, given that $(6,28,496)=(P_1,P_2,P_3)$ respectively:
$$\begin{align}6^2+5^2+2^6&=5^2+10^2 \\ 28^2+5^2+2^0&=27^2+9^2 \tag*{$\begin{pmatrix}(k,l)\,\,\,=&(6,10)&\text{resp.}\\ (k,l)\,\,\,=&(0,9)&\text{resp.} \\ (k,l)\,\,\,=&(3,32)&\text{resp.}\end{pmatrix}$}\\ 496^2+5^2+2^3&=495^2+32^2.\end{align}$$ This is as far as I have tested. I know that for every perfect number $P_n$, there exists a prime number $p$ such that $P_n=2^{p-1}(2^p-1)$ thus far (or at least, for every even perfect number), but this does not help me prove the conjecture. It is also conjectured that every perfect number is the sum of three cubes, but supposing this is valid does not help me prove this conjecture either.
And, writing the equation like the following does not help as well. $$2^k+(P_n+l)(P_n-l)=(P_n+4)(P_n+6).$$
Can it be verified whether or not my conjecture holds water?
Thank you in advance.