Can anyone help me prove this question? Thank you.
Prove that there exists a positive integer $n$ such that the number $2^n+3^n$ has exactly $23$ prime divisors.
(I have thought about some fomulas about $\omega (n)$, like ${\displaystyle \sum _{d\mid n}|\mu (d)|=2^{\omega (n)}}$, or construct the sequence ${n_k}$ such that $2^{n_{k+1}}+3^{n_{k+1}}$ has exactly one more prime divisor than $2^{n_{k}}+3^{n_{k}}$, but it seems to be hard.)
Proof by example:
$$2^{342}+3^{342}$$
Factors into exactly $23$ distinct prime factors:
$$ 13, 37, 61, 73, 181, 229, 2053, 8209, 13681, 15277, 25309, 102829, 30883969, 196498153, 724174057, 743780461, 2117021041, 2230888573, 54458107801, 8077765456081, 80381675102807053, 103911691734684541, 324469548901114381 $$