I need to find the prime factors of $2^{14}+3^{14}$ by hand (this was given in an exam at my university, so this is the motivation - I decided to state this because it may look unjustified to try to factor such a big number).
I began by noticing that $2^{14}+3^{14}=4^7+9^7=13\cdot 369181$ after applying a well known formula. So, it all boils down to finding the prime factors of $369181$.
Let $p$ be such a prime. Obviously, $p$ is not $2$ since our number is odd. We have $p|2^{14}+3^{14}$, so $2^{14} \equiv -3^{14} (p)$, where $(p)$ is shorthand for $\operatorname{mod} p$, so $2^{28}\equiv 3^{28}(p)$. Since $p$ is not equal to $2$, $2$ has an inverse modulo $p$, call it $2^{-1}$. Thus, $(3\cdot 2^{-1})^{14}\equiv -1(p)$ and $(3\cdot 2^{-1})^{28}\equiv 1(p)$. This easily implies that the order of $3\cdot 2^{-1}$ in the group $\mathbb{Z}_p^{\times}$ is $28$. So, $28$ divides $\varphi(p)=p-1$, that is $p=1+28t$ for some $t\in \mathbb{N}$.
This is where things got messy. I could only continue by taking the square root of $369181$, which is $607$ point something, so I have to check whether $369181$ is divisible by any prime of the form $28t+1$ that is less than $607$. These primes are $29, 113, 197, 281, 337, 421$ and $449$ if I didnt make any mistakes.
Now, I could do the computations, but they are really long. I wonder if there is some neat way to avoid doing this.
I believe you have done the best you can at avoiding tedious calculations. You have narrowed down the list of prime divisors to check to just $7$ primes. From here, you can check each of them quite easily by means of long division.
Another way to proceed, is to check whether $$(3\cdot2^{-1})^{14}\equiv-1\pmod{p},$$ for each of these primes. For the primes with $t$ even, this means $3\cdot2^{-1}$ is a quadratic residue mod $p$, and for primes with $t$ odd, this means $3\cdot2^{-1}$ is a quadratic nonresidue mod $p$. So it remains to determine $\left(\frac{6}{p}\right)$ for these primes.