We can generalize problem for any prime $p\equiv 2 (mod p)$ instead of $8$

Note that RHS is equal to (x+y+z)(sum of (x-y)^2)/2 Let's denote y=x+a,z=y+a+b, where a,b are natural numbers The ecuation becomes (3x+2a+b)(a^2+b^2+ab)=8^k Denote s(k)=the number of solutions for a given k=a(k)+b(k) where a(k) are the solutions in which a^2+b^2+ab is 1 and b(k) are the rest of the solutions Now if a^2+b^2+ab is 1 then {a,b}={0,1} so we obtain the ecuations 3x+2=8^k and 3x+1=8^k. Exactly one of those ecuations have a solution (depending on the parity of k), therefore a(k)=1 for any k If a^2+b^2+ab is even, then a and b are even, more than that, x is even too so we rewrite (3x'+2a'+b')(a'^2+b'^2+a'b')=8^(k-1), where x'=x/2,a'=a/2 and b'=b/2. Therefore b(k)=s(k-1) Since s(0)=1 we obtain that s(k)=k+1 for every k

Hint: $x+y+z=2^a, x^2+y^2+z^2-(xy+yz+zx)=2^b$, (or $3(xy+yz+zx)=2^{2a}-2^b$), which $a+b=3k$.

We prove by induction that for any nonnegative $k$ the answer is $k+1$: $$k=0:1=x^3+y^3+z^3-3xyz=(x+y+z)(x^2 +y^2 +z^2 -xy-xz-zy)\rightarrow (x,y,z)=(0,0,1)$$which is only solution. Assume for some $n-1$ we got $n$ solutions. Multiplying each of them by two we get $n$ solution for the case when $k=n$ and for any $(even,even,even)$ solution of case $k=n$ dividing each by two gives solution for $n-1$.And so we get $n$ even-even-even solution

We must proof that there is only one solution which one of them is odd .$$(x+y+z)=2^a;(x^2 +y^2 +z^2 -xy-xz-zy)=2^b;3|(a+b)$$$x+y+z>1$ So it is even and so two of them is odd and other is even but then second divisor must be odd or $$1=(x^2 +y^2 +z^2 -xy-xz-zy)=\frac{1}{2}((x-y)^2+(y-z)^2+(z-x)^2)\leftrightarrow z-x=1;y=z \text{ or } x$$if $x=y=a$ and $z=a+1$ then $$x^3+y^3+z^3-3xyz=3a+1$$and it is unique representation of $8^n =3a+1$