Cubic Cube?

First of all,what a beast!Congratulations on this problem,whether it's yours or not.My original solution solution is very long so i will skip through some parts of the solution hoping that someone could clarify it in the comments section.I'm really looking forward to the short solution you mentioned,although i have used pell's equation in the final step of my solution.Here we go!.It is easy to notice that $x\equiv2(mod 3)$ . By some trivial use of LTE from the condition we stated above,we can say that $v_3(x^3-2^3)=1+v_3(x-2)$ ,we have that $v_3(x-2)=y-1$ ,that is $x-2=3^{y-1} \cdot 7^{\alpha}$ .(A bit of clarification needed here.)Substituting into the original equation it is easy to see that $\alpha=0$ .So $x-2=3^{y-1}$ Substituting back into the original equation and dividing by $3^{y-1}$ ,we get the equation: $3^{2y-3}+6 \cdot 3^{y-2}+4=7^z$ From this it follows that $12 \cdot 7^z -12=m^2$ ,for some positive integer $m$ . Substitute $m=12q$ ,to get the equation $7^z-12q^2=1.$ .Reducing modulo 8,we arrive at the fact that $z=2z_1$ ,so we now have the pell equation $k^2-12q^2=1$ ,where $k=7^{z_1}$ .The fundemental solution is $k=7.q=2$ ,from where we get $x=11,y=3,z=2$ .Here comes the tricky part ,which for time reasons i will leave for others to prove.From the recurrence equation we have; $\begin{array} {lcl} x_{n+1} =7x_n+24y_n \\ y_{n+1} = 12x_n+7y_n \end{array}$ .It can now be proved by induction that the highest power of 7 which divides $x_n$ is 1,and the uniqueness of the solution now follows.Moderators should feel free to edit my solution so that it more comprehensible to other people,because its very late in my country.

$x^3-2^3=3^y*7^z~\implies~3|(x-2)(x^2+2x+4)~~and~~7|(x-2)(x^2+2x+4).\\ It~is~obvious~that~3|(x-2)~~and~~7|(x^2+2x+4).\\ Since~3^y*7^z+8~~~ is~odd, ~x ~has~to~be~odd.\\ So~x=5,11,....It~can ~not~be~5~~since~minimum~RHS=3^1*7^1+8=29.\\ Try~x=11.~~Then~(x^2+2x+4)=121+22+4=147=3*49.~~and ~~(x-2)=9.\\ \therefore~3^y*7^z=9*3*49=3^3*7^2.\\ \implies~~(x,y,z)=(11,3,2).~~\therefore~x+y+z=11+3+2=\Large~~\color{#D61F06}{16}.$

Uniqueness of solution has to be proved.