positive integer solutions to $x^3+y^3=3^z$ I am seeking all positive integer solutions to the equation $x^3+y^3=3^z$.
After doing number crunching, I think there are no solutions. But I am unable to prove it.
Attempt
If $x$ and $y$ have common divisor $d$, we have $d^3(m^3+n^3)=3^z$. So $d$ must be a power of $3$, and we are back to where we started. So we assume $x$ and $y$ are coprime.
Testing the parity, we have sum of 2 cubes to be odd. WLOG, we can assume $x$ is even and $y$ is odd.
Trying mod $3$, we have $x+y=0 \pmod 3$. Since $x$ and $y$ are coprime, $x$ and $y$ must be congruent to $1$ and $-1$ or vice-versa.
If I assume $x=3m+1$ and $y=3n-1$, expand out and simplify, I get $27(m^3+n^3)+27(m^2-n^2)+9(m+n)=3^z$. If I assume $z \geq 3$, this gives $(m^3+n^3)+(m^2-n^2)+\frac{m+n}{3}=3^{z-3}$. But I don't see how to proceed.
I also tried mod $9$ but didn't get anywhere, it didn't cut down the possibilities by much.
I also tried letting $y=x+r$. Then
\begin{align*}
x^3+y^3 &= x^3+(x+r)^3 \\
&= x^3 + (x^3+3x^2r+3xr^2+r^3) \\
&= 2x^3+3x^2r+3xr^2+r^3 \\
&= 3^z
\end{align*}
Then $3\mid 2x^3+3x^2r+3xr^2+r^3$, and $3\mid 3x^2r+3xr^2$, so this implies $3 \mid 2x^3+r^3$. But this doesn't yield any contradiction.
Can anyone supply a proof? Or if my hypothesis is wrong, how to derive all the integer solutions?
Thank you.

Solution by LTE, let $d=(x,y)>1$, $x=dx_1$, $y=dy_1$, $(x_1,y_q)=1$ then $$d^3\mid x^3+y^3=3^z$$ then $d\mid 3^a$ for a positive integer $a<z/3$. Dividing by $d^3$ we obtain $x^3_1+y^3_1=3^b$, for a positive integer $b=z-a$. This is a equation similar to the original, then we can assume $(x,y)=1$. If $3\nmid x+y$ then $x+y=1$ a contradiction. Then we have all the conditions to use LTE: $$v_3(x^3+y^3)=v_3(x+y)+v_3(3)=v_3(x+y)
+1=v_3(3^z)=z$$ From here we conclude that $x+y=3^{z-1}$ this implies that $x^2-xy+y^2=3$, $(x-y)^2+xy=3$, if $|x-y|\ge 2$ we have a contradiction, then there are two cases:
Case 1. $|x-y|=0$  this is not possible due $x$ and $y$ are coprimes and $x+y>2$.
Case 2. $|x-y|=1$ implying $xy=2$, with solutions $(x,y)=(2,1)$ or $(1,2)$, implying $3^{z-1}=2+1\Rightarrow z=2$.
Then the solutions to the original equation are: $(x,y,z)=(3^k,2\cdot 3^k,3k+2)$ or $(2\cdot 3^k, 3^k,3k+2)\forall k\in \mathbb{N}$.
