How Many Solutions 2

Since $20x^9 = 4269 - 117y^9$ is a multiple of $3$ , it follows that $x$ is a multiple of $3$ . But this implies that $4269 = 20x^9 + 117y^9$ is a multiple of $9$ , which is not the case. Thus there are no solutions.

$a^{18} \equiv 1 \pmod{19} \implies x^{9} \in \left\{-1, 0, 1 \right\}, y^{9} \in \left\{-1, 0, 1 \right\} \text{ when looking mod 19} \\ \text{ Looking mod 19 gives } x^{9} + 3y^{9} \equiv 13 \pmod{19} \text{ and } x^{9} + 3y^{9} \in \left\{-4, -3, -2, -1, 0, 1, 2, 3, 4 \right\} \\ 13\notin \left\{-4, -3, -2, -1, 0, 1, 2, 3, 4 \right\} \text{ so there are no integer solutions. }$

@William Allen Instead of looking $\operatorname{mod}19$ we can look $\operatorname{mod}9$ to make things easier........