Cubic diophantine equation with a prime $x^3 + y^3 + z^3 - 3xyz = p$. Question: Find all triple positive integers $(x, y, z)$ so that 
$$x^3 + y^3 + z^3 - 3xyz = p,$$
where $p$ is a prime number greater than $3$.

I have tried the following: The equation factors as
$$(x + y + z) (x^2 + y^2 + z^2-xy-yz-zx) = p.$$
Since $x + y + z> 1$, we must have $x + y + z = p$ and 
$$x^2 + y^2 + z^2-xy-yz - zx = 1.$$
The last equation is equivalent to 
$$(x-y)^2 + (y-z)^2 + (z-x)^2 = 2.$$
Without loss of generality you can assume that $x ≥ y ≥ z$, we have $xy ≥ 1$ and $xz ≥ 2$, implying 
$$(xy)^2 + (yz)^2 + (zx)^2 ≥ 6> 2.$$
Who can help me and correct me, thank you.
 A: Result: If $p>3$ is a prime number and  $x$, $y$ and $z$ are positive integers such that
$$x^3 + y^3 + z^3 - 3xyz = p,$$
then if $p\equiv1\pmod{3}$ we have, after permuting $x$, $y$ and $z$, that
$$(x,y,z)=\left(\tfrac{p-1}{3},\tfrac{p-1}{3},\tfrac{p+2}{3}\right),$$
and if $p\equiv2\pmod{3}$ we have, after permuting $x$, $y$ and $z$, that
$$(x,y,z)=\left(\tfrac{p+1}{3},\tfrac{p+1}{3},\tfrac{p-2}{3}\right).$$

Proof: As you already note, the equation can be expressed as
$$(x + y + z) (x^2 + y^2 + z^2-xy-yz-zx) = p,$$
which immediately shows that, because $x$, $y$ and $z$ must be positive,
$$x+y+z=p\qquad\text{ and }\qquad x^2 + y^2 + z^2-xy-yz-zx=1.\tag{1}$$
The latter can be rewritten as
$$(x-y)^2+(x-z)^2+(y-z)^2=2,$$
which shows that two of the three numbers are the same, and the third differs from them by only $1$. That is to say, without loss of generality we have
$$x=y=z\pm1.$$
Plugging this back into the first equation found at $(1)$ shows that
$$p=x+y+z=3x\pm1,$$
and so we find that $x=\tfrac{p\mp1}{3}$. As $x$ must be an integer we see that only one of the two choices of the $\pm$-sign is possible, depending on whether $p\equiv1\pmod{3}$ or $p\equiv2\pmod{3}$.
Conversely, a routine check shows that if $p\equiv\pm1\pmod{3}$ then the triplet of positive integers
$$(x,y,z)=\left(\tfrac{p\mp1}{3},\tfrac{p\mp1}{3},\tfrac{p\pm2}{3}\right),$$ and its three distinct permutations satisfy the equation
$$x^3 + y^3 + z^3 - 3xyz = p.$$
A: 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:
$$X=\frac{1}{6}(2a-3t\pm{k})$$
$$Y=\frac{1}{6}(2a+3t\pm{k})$$
$$Z=\frac{1}{3}(a\mp{k})$$
Thought the solution is determined by the equation Pell, but when calculating the sign was a mistake. There's no difference, but the amount should be. Therefore, the number of solutions of course.
I may be wrong, though. We still need to check other options.
