Suppose $a$ is a number of the form $8k\pm3$ (for example) and also suppose that there is some odd prime $p>3$ dividing $a$.
Claim: At least some $p|a$ has the property $p=8n\pm3$ for some $n\in\mathbb{Z}$.
To put it in different words, I want to show a special case of:
If $a$ is a number satisfying $a\equiv\pm b$ mod $c$ and $p|a$ for some odd prime >3, then we must have that $p\equiv\pm b$ mod $c$ for at least one $p$; + and - are free of choice in both $a$ and $p$, i.e. they don't need to match signs.
Little example: $a=35\equiv 3$ mod 8 and we have that $3<5|35$. Then $5\equiv-3$ mod 8. Hence, it is shown for one example.
The question still remains: is the statement in the yellow block true? Thank you for your time reading/answering this!
EDIT: A classmate gave me the proof idea for the $8k\pm3$ example.
Proof. Suppose $a\equiv3$ mod 8. All prime factors are either congruent $\pm1$ or $\pm3$ mod 8. Suppose $p_1,...,p_n|a$ with $p_i\equiv\pm1$ mod 8 $\forall i$. (Note that $i>1$ since otherwise $p_1=a$ where now all of a sudden $a\equiv\pm1$ mod 8 which is a contradiction to the definition of $a$.) Now factoring the $p_i$'s gives: $$a=\prod_{i=1}^{n}p_i=\prod_{i=1}^{n}(8k_i\pm1)=8l\pm1$$ where $l$ are all terms from the product that turn out to be factored with 64 or 16 from the product. Hence, we have a contradiction since $a\not\equiv1$ mod 8. Thus, we've proven that there has to be a prime of the form $8k\pm3$ in the decomposition of $a$.\hfill $\Box$.
Hence, the special case is proven. How about the general case?
Choosing only positive sign what you try to prove is that all integer of the form $f(x)=cx+b$ has a prime factor of the same form which seems a priori not true.
For example, take the form $f(x)=11x+5$ and put the value $a=f(2)=27=3^3$. This gives an easy counterexample and you have a lot of them giving the form of $f(x)$ above to all power $p^n$ of a prime.