What are the solutions to $a^2+ab+b^2$ $=$ $3^n$?

Question

What are all known integer solutions ($a, b, n$) to $a^2+ab+b^2$ $=$ $3^n$ besides ($1, 1, 1$) and ($-2, 1, 1$)? Do any others even exist?

This question comes from the identity that ($a^3±b^3$)/($a±b$) $=$ $0, 1$ $\pmod 3$. If ($a^3±b^3$)/($a±b$) $=$ $0$ $\pmod 3$, then let $3^n$ be highest power of $3$ dividing ($a^3±b^3$)/($a±b$). ($a^3±b^3$)/(($a±b$)$*3^n$) $=$ $1$ $\pmod 3$. When does ($a^3±b^3$)/(($a±b$)$*3^n$) $=$ $1$?

Answer 1

The set of solutions to $$ x^2 + xy + y^2 = 1 $$ in integers is finite (6).

 x = 1, y = 0 target 1
 x = -1, y = 0 target 1
 x = 1, y = -1 target 1
 x = -1, y = 1 target 1
 x = 0, y = 1 target 1
 x = 0, y = -1 target 1

The set of solutions to $$ x^2 + xy + y^2 = 3 $$ in integers is finite(6).

 x = 2, y = -1 target 3
 x = -2, y = 1 target 3
 x = 1, y = 1 target 3
 x = -1, y = -1 target 3
 x = 1, y = -2 target 3
 x = -1, y = 2 target 3

If $$ x^2 + xy + y^2 \equiv 0 \pmod 9 $$ then both $x,y$ are divisible by $3.$

This means that all solutions to your $3^n$ thing are $3^w$ times the items in the first two (finite) sets.

Answer 2

Let $a=u+v$ and $b=u-v$. It follows: $$a^2+ab+b^2=(u+v)^2+(u^2-v^2)+(u-v)^2$$ Hence $$3u^2+v^2=3^n$$ If $n=0$, then, $u=0$ and $v=\pm 1$

If $n=1$, then $v^2=3(1-u)(1+u)$ $\implies$ $(u,v)=(\pm 1,0),(\pm 1, \pm 1 )$

If $n>1$, then $u,v$ are both divisible by 3.