Set $x=y$... is there supposed to be an additional condition?

Thank professordad . Very sorry, I changed the problem. Because the problem is repeated(China Northern Mathematical Olympiad 2014 , Problem 3).

$n=10k+2$ I think does satisfy our answer. $11|3^{10k+2}+2$ and $3|5^{10k+2}+2$ The solution is easy to see and motivate.

In fact, consider any positive integer $m$. Denote $a=3^m + 2$ and $b = 5^m + 2$. Take $n = k\varphi(a)\varphi(b) + m$ for positive integers $k$. Clearly $\gcd(3,a) = \gcd(5,b) = 1$. Then $3^n + 2 \equiv 3^m + 2 = a \equiv 0 \pmod{a}$ and $5^n + 2 \equiv 5^m + 2 = b \equiv 0 \pmod{b}$.

Other answer is n=7•6^k where k some natural number.Lets check it: 3^(7•6^k)+2≡(-2)^(7•6^k)+2≡(mod11) by induction it is known that 3^n+2 always devided by 11 n is even.Therefore 5^n+2≡1+2≡0(mod3)