I only answer the (newly) added question (the others being adressed in comments):

Are there infinitely many primes of the form $a2^n + 1$, where $a$ is a fixed number?

Certainly *not* for each $a$. More precisely, Sierpiński (1060) showed that *there exist* infinitely many odd $a$ such that *all numbers* in the set
$$
\lbrace a2^n +1 \colon n \in \mathbb{N} \rbrace
$$
are *composite*.

Such an $a$ is called a Sierpiński Number; an explict example is $78557$. Chances are this is the smallest example, and there is some ongoing computing effort to show this. See the link I gave above for further details.

For certain other $a$ there are likely infinitely many, but this is never known. The point is that the most naive heuristic would be to say that the probability of $a2^n+1$ to be prime is proportional to $1/n$ (more precisely $1/\log (a2^n +1)$ by the Prime Number Theorem) and the series over $1/n$ being divergent one expects infinitely many, just like for Mersenne Primes.

However, and necessarily in view of what I said above, there can be problems with this heuristic: Depending on the $a$ there can be 'local' restrictions (that is one finds congruences that impede the number of this form to be prime, see again the site above).
Or/and, as in the Fermat case, there is a general factorization that reduces the range of the admissible exponents so much that the relevant series will converge and one thus expects at most finitely many.

One more related key-word: Primes of the form $a2^n + 1$ for $a$ odd not fixed, but $2^n \gt a$ (to avoid trivializing the condition) are called Proth primes.

Following the links on the two pages I gave you will find some more related notions and additional information.