All prime divisors $p$ of $2^{13}-1=8191$ have $p\equiv 1\mod 26$.
If $p$ divides $2^{13}-1$ then $2^{13}\equiv 1\mod p$, hence $2\in \Bbb F_p^\times$ has multiplicative order $13$. This gives us an order $13$ subgroup in $\Bbb F_p^\times$, hence by Lagrange's theorem $13$ divides $p-1$, the order of $\Bbb F_p^\times$. This says that $p\equiv 1 \mod 13$. Moreover, $p$ must be odd so we actually have $p\equiv 1 \mod 26$.
Question 1 Does that argument work?
Question 2 How do I conclude from this that $2^{13}-1$ is prime?
We showed that the prime divisors of that number are of the form $27, 53, 79, 102,\cdots$ and the first and last of those aren't prime, but I don't really see how to continue.
Best Answer
The brute force method to check the primality consists on dividing the number $n$ by every prime $\le\sqrt n$.
The argument you have used is correct, and narrows the search of possible prime divisors to those $p\equiv 1\pmod{26}$.
Since $\sqrt{8191}<91$ you have to check the divisibility by $53$ and $79$. Note that $27$ is not prime.