(1) The largest component $p^a$ of an odd perfect number is known to be greater than $10^{62}$.
(2) The largest prime of an odd perfect number is known to be greater than $10^8$.
Does (1) imply something stronger or weaker than (2)?
Let $LPF(n)$ denote the largest prime factor of $n$ and let $\sigma(n)$ denote the sum of the divisors of $n$.
I am interested in the minimum of $\max(p,LPF(\sigma(p^a)))$, where $p$ is prime and $p^a > 10^{62}$.
Not a complete or as organized as an answer should be, too long for a comment. (Also I suspect that most of what I write here is going to be stuff you already know.)
One naïve approach here is to try to do the following: Assume $N$ is an odd perfect number with $k$ distinct prime factors.
Given an odd perfect number N, we can bound $N$ from below in terms of its largest component. Then given N, we can bound $k$ in terms of $N$. If $p^a$ is a component of $N$, then $\sigma(p^a)p^a|2N$ since $\sigma(N)|N$, $(\sigma(p^a),p^a)=1$. Then we can use Pace Nielsen's result that $N < 2^{4^k}$. This gives a terrible bound, since we get from $(10^{62})^2 < 2N$, so $10^{124} < 2N$, and so one gets then from $N < 2^{4^k}$, that $k \geq 5$. But even a hundred or so years ago, $k \geq 6$ had been proven, and we know that $k \geq 10$ form other methods. And heck, if we go further and use the very extensive calculations you and Michael Rao did to get that $N$ is very big (you obviously know better than I do how far you've gone there, but last time I think we talked you had $N > 10^{2000}$), then combining that with $N < 2^{4^k}$, still only gets that $k \geq 7$ which is still well below the best bound we have now. Note that to get that the largest prime factor of $N$ was greater than $10^8$ this way, one would need a bound roughly around $k \geq 5 (10^6)$ which is sort of ridiculous.
There has been work on improving Nielsen's bound as mentioned, but the improvements are too small to make a difference here. I have in my notes a remark that Andrew Stone had a paper in preparation that was going to drastically improve on this type of bound, but it seems to be not out yet, and I don't have any preprint version? This note appears to be from about 5 years ago, so I don't know what is happening there.
But more broadly, none of the inequalities relating odd perfect numbers seem to be remotely strong enough that if one has three functions of arithmetic properties $N$, say $a(N)$, $b(n)$ and $c(n)$, that if we any non-trivial lower bounds for $c(n)$ that those bounds are almost ever exceeded by a path from some lower bound for $a(n)$ to $c(n)$ through bounds for $b(n)$.
So any sort of bound here that has any hope of being non-trivial is going to as you implicitly observed depend on bounding the largest prime factor of $\sigma(p^a)= \frac{p^{a+1}-1}{p-1}$. But even this looks difficult. The ABC conjecture with plausible constants can rule out silly things like very big solutions of $\frac{p^{a+1}-1}{p-1} = 3q^b$ where $a$ and $b$ are large. But there does not seem to be anything which rules out $\frac{p^{a+1}-1}{p-1}$ having many distinct small or medium sized prime factors. And the very worst case scenario is going to be where $a=1$, in which case we care about $\sigma(p)=p+1$. In that situation, we could even have $p+1 = 2 (3^m)$ for some $m$. Now, it should still be the case that $\sigma(3^d)$ for some $d \geq m$ would still at least one moderately big prime factor, but proving that seems difficult, and if one had instead $p+1$ is a product of 2 times a bunch of moderately sized powers of primes, the situation looks even worse (and then there are also a lot more such primes). Now, in this situation, we don't directly care about this situation, since if $a=1$, then you get that the largest prime factor is at least $10^{62}$ for free. So one should hope that with small $a$, one doesn't have similar situations, but ruling that out seems difficult. Slightly more generally, if one has $a \leq 6$ then one gets a better bound than $10^8$, (actually also for $a=7$ but no odd perfect number has $p^7||n$), so one would hope that some sort of argument involving that might be doable, but it doesn't jump out in any obvious way.
Most likely the most effective way of improving the $10^8$ bound of Ohno and Goto is just pushing the computation further, and since their paper is from 2008, improvements in algorithms and computations probably mean that improving that should be viable.
While commenting on this, I'll note that there are a pair of related papers by Doug Iannucci from about 20 years ago where he proved lower bounds on the second and third largest prime factors of $N$. If the second largest prime factor is $p_{k-1}$ and the third largest is $p_{k-2}$ then he showed that $p_{k-1} \geq 10^4$ and that $p_{k-2} \geq 100$. For years I've had on my to-do list ending those computations. There are two reasons that this seems like a plausible thing to aim for, aside from the obvious fact that there is just a lot more computational power available right now: First, in general we have about 20 years worth more understanding of odd perfect numbers and cyclotomic polynomials to leverage. Second, and more narrowly, Doug's results use heavily the earlier bound of Cohen and Hagis that the largest prime factor of $N$ must be at least $10^6$. But as you noted we have Goto and Ohno's much more extensive computation now that the largest prime factor must be $10^8$, and so the Iannucci type approach has more room.
I have specific ideas about how to improve some aspects of his approach. However, the Iannucci papers are tough to follow, with a lot of primes eliminated in orders where the reason those orders are chosen is opaque, and even having read the papers multiple times, there are bits I don't have a really good understanding of. Unfortunately, the obvious person to ask about this was Doug, and he passed away in 2020. So this last part of this remark is essentially a project proposal for you or anyone else who wants to work on that aspect of the problem, I'd be happy to help out.
