Finding all natural $n$ such that $2^n+2^{2n} +2^{3n}$ has only $2$ prime factors. Find all natural $n$ such that $2^n+2^{2n} +2^{3n}$ has only $2$ prime factors. 
I've tried checking the first 6-7 $n$'s on wolframalpha, but I don't see any patterns for even nor odd $n$'s. At first I thought for all odd $n$'s it was divisible by $2,3,7$, but $n=5$ doesn't work ($n=1,3,7$ do work). How are these $n$ found?

This is just a sketch of a partial answer...
$$2^n+2^{2n}+2^{3n}=2^n(1+2^n+2^{2n})$$
so you need to check if $1+2^n+2^{2n}$ is of the form $p^k$. First note that, if $n$ is even
$$1+2^n+2^{2n}\equiv 1 + 1 + 1\equiv 0\bmod 3$$
so, if $n$ is even, you need
$$2^n+2^{2n}=3^a-1$$
if $a$ is odd, then $3^a-1\equiv -1-1\equiv -2\bmod 4$, so $a$ is even. Then, by LTE
$$v_2(3^a-1)=v_2(3+1)+v_2(3-1)+v_2(a)-1=2+v_2(a)$$
but also $v_2(2^n+2^{2n})=n$ so $n=2+v_2(a)$, hence $2^{n-2}\vert a$, but then
$$2^n+2^{2n}\leq 3^{2n+1} < 3^{2^{n-2}}$$
for $n\geq 5$. So no solutions for $n$ even greater than $5$.
Moreover, if $3\not\vert n$, then $2^n\equiv 2,4\bmod 7$, so $1+2^n+2^{2n}\equiv 0\bmod 7$. The same reasoning as before says that $1+2^n+2^{2n}$ is not a power of $7$ if $n\geq 5$ and coprime with $3$.
We are left with odd multiples of $3$. We see that for $n=3$ we get $73$, which is prime; repeating the previous reasoning, we exclude the numbers of the form $9k+3$ and $9k+6$, but we have still to check odd multiples of $9$.
This strategy doesn't seem promising, because $9$ is a solution, $27$ is not but in the factorization of the number we obtain with $n=27$ other primes ($4$-digit primes!) show up...
