There are several ways to solve this problem. One of them uses the binomial expansion. $$\begin{align*} &3^{4n+2} + 5^{2n+1} = 9^{2n+1} + 5^{2n+1} = (14-5)^{2n+1} + 5^{2n+1}\\ \\=\ & \left(\binom{2n+1}{0}\cdot 14^{2n+1} - \binom{2n+1}{1}\cdot 14^{2n}\cdot 5 +\cdots \right. \\ +& \left. \ \binom{2n+1}{2n}\cdot 14\cdot 5^{2n} - \binom{2n+1}{2n+1} \cdot 5^{2n+1}\right) + 5^{2n+1}\\ \\ =\ &\binom{2n+1}{0}\cdot 14^{2n+1} - \binom{2n+1}{1}\cdot 14^{2n}\cdot 5 + \cdots + \binom{2n+1}{2n}\cdot 14\cdot 5^{2n}\end{align*}$$

Because every term is a multiple of $14$, therefore their sum, i.e. $\left(3^{4n+2} + 5^{2n+1}\right)$, must be divisible by $14$.

This can also be expressed in a more concise way using polynomial factorization: $$9^{2n+1}+5^{2n+1} = (9+5)\times(\cdots) = 14\times (\cdots)$$