An elaboration of a step in an example. A part of the example is given below:

Example 9.7. Let us turn to a quite different application of these ideas. At an earlier stage, it was observed that if $F_n = 2^{2^n} + 1$, $n > 1$, is a prime, then $2$ is not a primitive root of $F_n$. We now possess the means to show that the integer $3$ serves as a primitive root of any prime of this type.
As a first step in this direction, note that any $F_n$ is of the form $12 k + 5$. A simple induction argument confirms that $4^m \equiv 4 \pmod {12}$ for $m = 1, 2, \ldots$; hence, we must have
$$F_n = 2^{2^n} + 1 = 2^{2 m} + 1 = 4^m + 1 \equiv 5 \pmod {12}$$

But I do not understand why $2^{2^{n}} = 2^{2m}$? could anyone explain this for me please?
Is the above mentioned $F_{n}$ is Fermat numbers? if so why the $n$ starts from 2 here?
 A: 
But I do not understand why $2^{2^{n}} = 2^{2m}$?

At this point in the text, the author has defined $m$ as $2^{n-1}$. (The author hasn't written this definition explicitly; presumably, they assumed that the reader would be able to figure it out easily.) Since $m = 2^{n-1}$,
$$2^{2^{n}} = 2^{(2 \cdot 2^{n-1})} = 2^{2 m}.$$

Is the above mentioned $F_{n}$ is Fermat numbers? if so why the $n$ starts from 2 here?

Yes, $F_n$ is the Fermat numbers here. The author writes that for $n > 1$, if $F_n$ is prime, then $2$ is not a primitive root of $F_n$. Here, $n$ must be greater than $1$ because $2$ is a primitive root of $F_1$ (which is $5$).

what about the $m$ beside the congruence at the third line from below? are there 2 different m's?

Yes, those are two different $m$'s. The author could just as well have used a different letter for one or the other, or both. However, the author intentionally used the same letter in both places, in order to indicate that the two variables "match up," so to speak.
