Let $n$ be a positive integer. What is the number of solutions to the equation $$8^n = a^3+b^3+c^3-3abc$$ with integers $a\geq b\geq c\geq 0$?
We have the factoring $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$ meaning that both factors must be powers of $2$.
Note $$a^2+b^2+c^2-ab-ac-bc=(1/2)((a-b)^2+(a-c)^2+(b-c)^2)$$ Now suppose $a+b+c$ is even. Then all three variables are even, or else exactly one of them is even. If all three are even, then we can divide the original equation through by 8, getting a smaller solution. Repeat this enough times, and we get to a solution with exactly one even variable; let it be $a$. Then $a^2+b^2+c^2-ab-ac-bc$ is odd, but it's a power of 2, so it must be 1, so $(a-b)^2+(a-c)^2+(b-c)^2=2$. The only way to get three squares to sum to $2$ is for one of them to be zero and the others to be $1$, so $a-b=\pm1$, $a-c=\pm1$, $b=c$. Taking the plus sign, we get $3a-2=2^r$ for some $r$, so $a=(2^r+2)/3$, $b=c=(2^r-1)/3$. Then $$a^3+b^3+c^3-3abc=(1/27)((2^r+2)^3+2(2^r-1)^3-3(2^r+2)(2^r-1)^2)=2^r$$ so we need to take $r$ to be a multiple of $3$. For example, $r=6$ leads to $a=22$, $b=c=21$, $a^3+b^3+c^3-3abc=64=8^2$.
The reader may enjoy working through the case with the minus sign.
Let $\zeta $ be the principal third root of unity. Then $$\left ( a+b+c \right ) \left ( a+b \zeta +c \zeta ^{2} \right ) \left ( a+b \zeta ^{2} +c \zeta ^{4} \right ) = a ^{3} + b ^{3} + c ^{3} - 3abc$$ (Similar formulae exist for more variables, but they don't look nearly as nice).
If $8^{n}$ is a solution to your equation, then:
$$\left ( a+b+c \right ) \left ( a+b \zeta +c \zeta ^{2} \right ) \left ( a+b \zeta ^{2} +c \zeta ^{4} \right ) = 8^{n}$$
Now observe, crucially, that $a+b+c$, $a+b \zeta +c \zeta ^{2}$ and $a+b \zeta ^{2} +c \zeta ^{4}$ form a basis in $\mathbb{C}$. So, ignoring the RHS, we can obtain any three multiplicants on the LHS.
Furthermore, we are currently working over the Eisenstein integers $\mathbb{Z}\left [ \zeta \right ]$, which is a UFD. Therefore if we factorize $8^{n}$ over $\mathbb{Z}\left [ \zeta \right ]$ on the RHS, we must pair up these factors on the LHS over $\mathbb{Z}\left [ \zeta \right ]$.
To illustrate, the unique Eisenstein prime factorization in the ring-theoretic sense is $8^{n}=2^{3n}$. If we group its factors in some way, we obtain $a,b,c$. For instance, $8^{n}=2^{n}2^{n}2^{n}$ gives $a=2^{n}$, $b=c=0$ from: $$a+b+c=2^{n}$$ $$a+b \zeta +c \zeta ^{2}=2^{n}$$ $$a+b \zeta ^{2} +c \zeta ^{4}=2^{n}$$
But that's not the end of it. Any given factor may be multiplied by any one of $\left \{ \pm 1, \pm \zeta , \pm \zeta ^{2} \right \}$ (because we are allowed to multiply primes by units). For instance, another solution may come from multiplication of the first two factors by $-1$: $$a+b+c=-2^{n}$$ $$a+b \zeta +c \zeta ^{2}=-2^{n}$$ $$a+b \zeta ^{2} +c \zeta ^{4}=2^{n}$$
Which gives $a=-2^{n}$, but we don't want negative solutions. Multiplication of the first two by $\zeta $ gives $a=\zeta 2^{n}, b=c=0$ and so on and so forth. Many additional solutions can be discarded.
Let us try this instead: $8^{n}=\left ( 2^{n} \right ) \left ( 2^{n} \zeta \right ) \left ( 2^{n} \zeta^{2} \right )$. This gives: $$a+b+c=2^{n}$$ $$a+b \zeta +c \zeta ^{2}=2^{n} \zeta$$ $$a+b \zeta ^{2} +c \zeta ^{4}=2^{n} \zeta^{2}$$
And then $b=2^{n}, a=c=0$.
This illustrates that any factorization gives rise to a number of different ones. It is not hard to see that given a factorization $\left ( a,b,c \right )$, the only other meaningful solutions are permutations, as we expect. Not much has been found in this regard, but it was worth checking.
With that intuition, let's start counting. Starting with: $$8^{n}=ABC$$ For $A,B,C$ in $\mathbb{N}$, we have the corresponding system of equations: $$a+b+c=A$$ $$a+b \zeta +c \zeta ^{2}=B$$ $$a+b \zeta ^{2} +c \zeta ^{4}=C$$
We obtain: $$3a=A+B+C$$ $$3b=A+B\zeta^{2}+C\zeta$$ $$3c=A+B\zeta+C\zeta^{2}$$
We note that $A,B,C$ are all powers of $2$. These powers mod 3 give rise to the alternating sequence $\left \{ +1,-1,+1,-1\cdots \right \}$, so that $A+B+C$ is only divisible by $3$ iff either $A,B,C$ are all even or all odd powers of $2$ (i.e. exponents have the same parity).
Now we also require $B\zeta^{2}+C\zeta$ is real. This means, taking complex conjugates (noting the conjugate of $\zeta$ is $\zeta^{2}$ and vice versa) that:
$$\left ( B\zeta ^{2} + C\zeta \right ) = \overline{ \left ( B\zeta ^{2} + C\zeta \right )}$$ If and only if
$$B=C$$
Then our system simplifies to: $$3a=A+B+C$$ $$3b=3c=A-B$$
And that is all we need. To illustrate, take $n=2$, i.e. $8^{2}=AB^{2}$. We have only the choices $A=2^{6}$ and $B=2^{0}$, $A=2^{2}$ and $B=2^{2}$. The first corresponds to $a=22, b=c=21$, the second to $a=4, b=c=0$. Those are the only solutions.
Take $n=5$, then $8^{5}=2^{15}=AB^{2}$, we have the choices $A=2^{13}$ and $B=2^{1}$, $A=2^{9}$ and $B=2^{3}$, $A=2^{5}$ and $B=2^{5}$ and those are the only solutions.
Hard one: take $n=6$, then we have the choices:
$$\left ( A,B \right ) \in \left \{ \left ( 2^{18}, 2^{0} \right ) \left ( 2^{14}, 2^{2} \right ) \left ( 2^{10}, 2^{4} \right ) \left ( 2^{6}, 2^{6} \right ) \right \}$$
This gives: $$8^6=87382^{3}+87381^{3}+87381^{3}-3\cdot 87382 \cdot 87381 \cdot 87381$$ $$8^6=5464^{3}+5460^{3}+5460^{3}-3\cdot 5464 \cdot 5460 \cdot 5460 $$ $$8^6=352^{3}+336^{3}+336^{3}-3\cdot 352\cdot 336 \cdot 336 $$ $$8^6=64^{3}+0^{3}+0^{3}-3\cdot 64\cdot 0\cdot 0 $$
It's easy to see now that the answer to your question is precisely:
$\frac{n+2}{2}$ if $n$ is even.
$\frac{n+1}{2}$ if $n$ is odd.
Strangely enough, the solution is finite.
for the equation:
$$X^3+Y^3+Z^3-3XYZ=q=ab$$
If it is possible to decompose the coefficient as follows:
$4b=k^2+3t^2$
Then the solutions are of the form:
$$X=\frac{1}{6}(2a-3t\pm{k})$$
$$Y=\frac{1}{6}(2a+3t\pm{k})$$
$$Z=\frac{1}{3}(a\mp{k})$$
