(Algebraic manipulations, using the fact that if $a \mid b$, then $a \mid b+k a$ for any integer $k$)

Let $n \geq 1$ such that $3^{n-1}+5^{n-1} \mid 3^{n}+5^{n}$. Then

$$
3^{n-1}+5^{n-1} \mid 5\left(3^{n-1}+5^{n-1}\right)-3^{n}+5^{n}=2 \cdot 3^{n-1}
$$

But $3^{n-1}+5^{n-1}>2 \cdot 3^{n-1}$ for $n \geq 2$, so $3^{n-1}+5^{n-1}$ does not divide $2 \cdot 3^{n-1}$ for $n \geq 2$. Conversely, we verify that $n=1$ works.
Find all integers $n \geq 1$ such that $3^{n-1}+5^{n-1}$ divides $3^{n}+5^{n}$.