I just heard a This American Life episode which recounted the famous anecdote about Frank Nelson Cole factoring $N:=2^{67}-1$ as $193{,}707{,}721\times 761{,}838{,}257{,}287$. There doesn't seem to be a historical record of how Cole achieved this; all we have I could find his statement that it took "three years of Sundays". Ira Glass's guest, Paul Hoffman, suggests that this was done by trial division.

But this is nuts, unless I am missing something. Three years of Sundays is $156$ days. If he works $10$ hours a day, that's $93{,}600$ minutes. There are $10{,}749{,}692$ primes up to $193{,}707{,}721$. So that is more than $100$ trial divisions a minute. Worse than that, existing prime tables didn't go high enough: According to Chapter XIII of Dickson's History of the Theory of Numbers, existing tables of primes only ran to something like $10{,}000{,}000$ ($664{,}579$ primes), so for the vast majority of the trial divisions, he'd have to find the primes first. (Lehmer, in 1914, went up to $10{,}006{,}721$.)

But I'm puzzled thinking what else Cole could have done. I skimmed Chapter XIV in Dickson. The methods which seem to have existed at the time are:

Various ways to speed up trial division for the first $1000$'s of prime numbers. That only helps at the start.

Writing $N$ as $x^2-y^2$. But $y$ would be $380{,}822{,}274{,}783$, which is an even larger search.

Since $2$ is a square modulo $N$ (namely, $(2^{34})^2 \equiv 2 \mod N$), we know that all prime factors must be $\pm 1 \bmod 8$, which cuts the time in half. But that's only a factor of $2$.

Since $N \equiv 3 \mod 4$, there must be a prime factor which is $3 \bmod 4$, so we could try only checking those primes. But this turns out to make things worse, since the SMALL factor is the one which is $1 \bmod 4$.

If we could write $N$ as a sum of squares in two ways, we'd be done. But $N$ isn't a sum of squares, since it is $3 \bmod 4$. Generalizations to other positive definite quadratic forms were known at the time, but how would Cole know which quadratic form to try?

A variant of the above would be to use the quadratic form $2x^2-y^2$, since we already have one solution. Dickson doesn't mention any work using mixed signature forms, but it would work. And since $\mathbb{Z}[\sqrt{2}]$ is a PID, there must be a second way to write $N$ as $2x^2-y^2$, not related to the previous by units of $\mathbb{Z}[\sqrt{2}]$. I'm not sure how large this second solution is.

So, my question is:

How could someone find the prime factors of $N$ in $100{,}000$ minutes of hand computation?