Find all natural number $n$ for which $3^9+3^{12}+3^{15}+3^n$ is a perfect cube. 
Find all natural number $n$ for which $3^9+3^{12}+3^{15}+3^n$ is a perfect cube.

What I have tried.
$3^9+3^{12}+3^{15}+3^n$
$=3^9(757+3^{n-9})$
Let $757+3^{n-9}=a^3$
Taking modulo $3$:
$a^3\equiv 1 \pmod 3$
$\implies a\equiv 1 \pmod 3$
Also, $a^3>757>729=9^3$
$\therefore a =10+3k$
$a=10$ gives $n=14$ 
Now, how can I know if there is any other $n>14$ satisfying the given condition.
PS: Please not use computer programmes to answer. I want pure mathematical solution.

Check manually if there exist any solution with $n \le 9$.
Now assume $n>9$, and look at the equation $3^m+757=a^3$ modulo 7 (with $m=n-9$):
$3^m+1 \equiv 3^m+757 \equiv a^3 \equiv \{0,-1,1\} \pmod 7$
$3^m \equiv \{0,-1,-2\} \pmod 7$
$n \equiv \{3,5\} \pmod 6$
If $m=6k+3$,
$ 757 = a^3-(3^{2k+1})^3 = (a-b)(a^2+ab+b^2)$
where $b=3^{2k+1}$. Factoring 757, you can check there is no solutions.
Else, if $m=6k+5$,
$757 = a^3-3^5(3^{2k})^3 = a^3-3^5b^3$
where $b=3^{2k}$. This is a Thue equation, effectively solvable:
using PARI/GP

tnf = thueinit(x^3-243)
thue(tnf, 757)
[[10, 1]]

you can check the only solution is $(a,b)=(10,1)$, hence $(a,n)=(10,14)$
