Number of distinct near-squares primes dividing an odd perfect number I'm curious about if the following question is in the literature or what work can be done about it.
Denote the number of distinct primes dividing an odd perfect number $N$ with the arithmetic function $\omega(N)$. Wikipedia has this section dedicated to near-square primes from the article Landau's problems.
Under the assumption that there exists an odd perfect number $N$, we denote the number of distinct  near-squares primes dividing $N$ as $\omega_{\text{nsp}}(N)$, thus $\omega_{\text{nsp}}(N)\leq \omega(N)$.
Here I add as general reference that a near-square prime is a prime number of the form $p=n^2+1$, that's the sequence A002496 from The On-Line Encyclopedia of Integer Sequences.

Question. I would like to know if we can improve the previous lower bound and/or the upper bound, in this way $$\text{lower bound}<\omega_{\text{nsp}}(N)<\text{upper bound}\tag{1}$$
being these bounds functions of $N$, that's $\text{lower bound}=\text{lower bound}(N)$ and $\text{upper bound}$ denotes a function $\text{upper bound}(N)$. Many thanks.

I can to deduce obvious improvements for particular cases of Touchard's theorem, or when I consider Euler's theorem for odd perfect numbers. Also I know bounds for the radical of an odd perfect number, bounds for the Euler's totient function $\varphi(N)$, and I know the product formula representation for these arithmetic functions.
 A: In general, very few prime factors in an odd perfect number can be of the form $n^2+1$.
In particular, if N is an odd perfect number then $\frac{\sigma(N)}{N}=2$, and for any $m$ (perfect or not), $\frac{\sigma(n)}{n} \leq \prod_{p|n}\frac{p}{p-1}$ with equality if and only if $n=1$.
Thus, for any perfect number, one needs $ \prod_{p|N}\frac{p}{p-1}>2$. However,
$$\prod_{p, p =i^2+1}\frac{p}{p-1} <   \prod_{i=1}^{\infty} \frac{(2i)^2+1}{(2i)^2}  < 1.4$$
So, only a small fraction of the primes can be of the form $n^2+1$. Turning this into a bound on $\omega(N)$ directly is going to be very difficult, because one cannot rule out that one has an odd perfect number with a few primes not of this form, and then a lot of very big primes of the form $n^2+1$ that contribute very little to the product.
Note that similar remarks would apply to almost any class of primes of the form $p=Q(n)$ for some polynomial $Q$. There's very little content here dealing with the fact that we're looking at $n^2+1$ at all.
