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.
 A: 
Not a full solution, I am simply reducing the problem to that of listing integer points on two elliptic curves. IIRC this is implemented in some dedicated CAS, and therefore this gives us a route to a definite answer.

With small values of $n$ checked by brute force, we can cancel the factor $3^9$. We are thus left with the equation
$$
1+3^3+3^6+3^{n-9}=x^3\Longleftrightarrow 757+3^{n-9}=x^3.\qquad(*)
$$
Depending on the residue class of $n$ modulo three we can write $3^{n-9}=3^\epsilon y^3$ with $\epsilon\in\{0,1,2\}$.
This means that any integer solution of $(*)$ will give rise to an integer
solution of one of the following Diophantine equations
$$
\begin{aligned}
x^3&=y^3+757,\\
x^3&=3y^3+757,\\
x^3&=9y^3+757.
\end{aligned}
$$
Each of these defines an elliptic curve. Those are known to have only finitely many integer points $(x,y)$, and (IIRC) algorithms for finding them exist (and are available in CAS's heavily used by number theorists).

Given such finite lists, we can quickly check whether $y$ can be a power of three in any of them.

Further remarks:


*

*The first elliptic curve won't produce solutions. We have
$$y^3<y^3+757<(y+1)^3$$ whenever $y>16$, and it is easy to check that the powers of three in this range won't give us any solutions.

*Peter did an extensive computer verification for a largish range of values of $n$ (see the comments under main). So even an upper bound on the integer points (don't remember whether useful ones are known) will help us settle the main question

A: 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 $(a,n)=(10,14)$
A: Only long comment.
Let $n-9=x$, then we have equation $757+3^x=a^3$ for positive integers $a,x$.
$2\mid a$ and $a\equiv 1\pmod3$, then some prime $2<p\mid a$ and $p\equiv 2\pmod3$.
Find suitable primes $p$ and related $x$ using descrete logarithm, multiplicative order and CRT in pari/gp.
First, $x$ is odd, becose $8\mid a^3$, $ord_8(3)=2$ and $zlog_{3_{8}}(-757)=1$:
? m=Mod(3,8);
?
? h=znorder(m)
%1 = 2
?
? znlog(-757,m,h)
%2 = 1

gp-code for finding $p$ and $x$:
 forprime(p=5,100, if(p%3==2,
  m=Mod(3,p);
  h=znorder(m);
  z=znlog(-757,m,h);
  if(z,
   c=iferr(chinese(Mod(1,2),Mod(z,h)), E, 0);
   if(c,
    m=Mod(3,p^2);
    h=znorder(m);
    z=znlog(-757,m,h);
    if(z,
     c=chinese(c,Mod(z,h));
     m=Mod(3,p^3);
     h=znorder(m);
     z=znlog(-757,m,h);
     if(z,
      c=chinese(c,Mod(z,h));
      j=c.mod; c=lift(c);
      print("p = "p"    x = "c" + k*"j);
     )
    )
   )
  )
 ))

Output:
p = 5    x = 5 + k*100
p = 23    x = 8169 + k*11638
p = 29    x = 11719 + k*23548
p = 47    x = 28121 + k*101614
p = 71    x = 332919 + k*352870

I.e. if $23\mid a$, then $x=8169 + k\cdot 11638$, where $k$ is 0,1,2,....
