Find the integer solution for $3^x+27^y+81^z=10935$

This is a rather thickish question so we have to be very careful in our simplifications.
The left hand side of the equation can be expressed in base $3$, so we must look for a way to express the right hand side in base $3$ also.
Let's take a walk.

👉 $3^x+3^{3y}+3^{4z}=2187\times{5}$
$3^x+3^{3y}+3^{4z}=3^7\times{5}$
Factor out $3^x$ from the left hand side 
$3^x(1+3^{3y-x}+3^{4z-x})=3^7(5)$
We can now equate the right hand side with the left hand side.
👉 $3^x=3^7$
👉$x=7$
And
$1+3^{3y-x}+3^{4z-x}=5$
Collect like terms
$3^{3y-x}+3^{4z-x}=4$
Both sides of the equation should be expressed in base 3 so that we can find $y$ and $z$.
$3^{3y-x}+3^{4z-x}=3+1$
$3^{3y-x}+3^{4z-x}=3^1+3^0$
There are two cases to finding $y$ and $z$
👉 CASE I
$3^{3y-x}=3^1$
$3y-x=1$
Substitute $x=7$
$y=\frac{8}{3}\notin\mathbb{Z}$
And
$3^{4z-x}=3^0$
$4z-x=0$
Substitute $x=7$
$z=\frac{7}{4}\notin\mathbb{Z}$

👉CASE II
$3^{3y-x}=3^0$
$3y-x=0$
Substitute $x=7$
$y=\frac{7}{3}\notin\mathbb{Z}$
And 
$3^{4z-x}=3^1$
$4z-x=1$
Substitute $x=7$
$z=\frac{8}{4}=2\in\mathbb{Z}$

Hence, there are no  integer solution for $3^x+27^y+81^z=10935$ since only $x=7$ and $z=2$ are integers.

You might use a different approach to solve the question, if you are getting a different answer or you feel my method is inappropriate, kindly drop a comment/observation in the comment section. Thank you.