Here is the hint to the proof given in the book:
Using the following 2 theorems:
1-If $p$ is an odd prime, then any prime divisor of $M_{p}$ is of the form $2kp +1$.
2-If $p$ is an odd prime, then any divisor $q$ of $M_{p}$ is of the form $q \equiv \pm 1 \pmod 8 .$
and the hint said that then the only primes you have to test are 191, 457 and 647.
But I do not understand how those numbers are calculated, could anyone explain this for me please?
So first get an estimate of size (the following is intended to be achievable without a calculator or table of primes):
$$M_{19}=2^{19}-1=2^{10}\cdot2^9-1=1024\cdot 512-1=524287\lt 725^2$$
Whence if $M_{19}$ is not a prime it will have a least one prime factor less than $725$
Then the first theorem tells us that possible factors are numbers of the form $38k+1$(if they are prime), so $39,77,115,153 \dots$ and we note that all of these early ones have prime factors which are easy to spot. Every third one of these from the first will be divisible by $3$, every fifth from the third by $5$, every seventh from the second by $7$ etc.
Since $38\times 20 \gt 725$ we start with a list of $19$ values for k, with obvious factors:
$k=1: 3\times 13$ (excludes $k=4,7,10,13,16,19; k=14$)
$k=2: 7\times 11$ (excludes $k=9,16;13$)
$k=3: 5$ (excludes $k=8,13,18$)
This leaves $k=5, 6, 11, 12, 15, 17$ as possibilities.
Since $38\equiv 6 \bmod 8$ these give congruences $-1, 5,3,1,3-1$ modulo $8$ and only $k=5,12,17$ need be tested. Note: these three are not yet known to give primes, but have no prime factor $3,5,7,11,13$ since these were excluded at the first stage. [If you want to be more analytical the mod $8$ congruence excludes $k\equiv 2,3 \bmod 4$]