What are the solutions in numbers of $xyz | x^n + y^n + z^n$, $n$ prime. What are the integers $(x,y,z)\in \mathbb Z^*$ such that $xyz$ divide $x^n + y^n + z^n$, for prime $n$?
I have no other motivation for that problem but its inherent beauty and interest.
Note that it can be assumed without loss of generality that $x\geq |y|\geq |z|$.
Here is what I've obtained so far:
If $xyz$ divides $x+y+z$ and $n$ is odd, $xyz$ divides $x^n + y^n + z^n$. Since the set of solution $(x,y,z)$ of the former relation for which $x\geq |y|\geq |z|$ is
$$\{(2,1,1), (1,1,1), (3,2,1), (x,1,-1), (x,-1,1):\ x\in \mathbb N^*\} \cup \{(x,y,z): x+y+z = 0\},$$
these are also solutions of the proposed equation.
The proof by induction of the above proposition is based on the following formula:
$$x^n+y^n+z^n = (x^{n-1}+y^{n-1}+z^{n-1})(x+y+z) - (x^{n-2}+y^{n-2}+z^{n-2})(xy+xz+yz) + (x^{n-3}+y^{n-3}+z^{n-3})xyz.$$
Regarding the set of solutions of the equation $xyz|x+y+z$, this is not entirely trivial but is nevertheless straightforward.
So, let me call the above solutions the "straightforward solutions".
My question is: do there exist solutions that are not straightforward?
I guess that the case $n=2$, not dealt with in the above proposition, is very different from the case "$n$ odd".
EDIT: to avoid solutions by scaling (see the answer of Qiaochu Yuan), I should probably have added the condition $\gcd (x,y,z) = 1$. I suggest to restrict the problem to that case from now on.
That is, the question which is the closest to what I intended is:
do there exist globally coprime solutions that are not straightforward?
 A: (This is an answer to your body question but nowhere close to an answer to your title question. In general it's bad practice to ask different questions in the title and the body.)
I don't see a reason to restrict to prime $n$ so let's consider all $n$. If we substitute $(x, y, z) = c(r, s, t)$ for some scale factor $c$ we get that we want $c^3 rst$ to divide $c^n (r^n + s^n + t^n)$. So the problem changes character quite a bit depending on whether $n \le 2, n = 3$, or $n \ge 4$. For $n \ge 4$ scaling gives that we want $rst$ to divide $c^{n-3}(r^n + s^n + t^n)$ so now we get solutions just by setting $c = rst$; that is, for $n \ge 4$ we get a $3$-parameter family of solutions
$$(x, y, z) = rst (r, s, t), r, s, t \in \mathbb{Z} \setminus \{ 0 \}.$$
($\mathbb{Z}^{\ast}$ is not correct notation for $\mathbb{Z} \setminus \{ 0 \}$; this means the group of units which is much smaller, just $\{ \pm 1 \}$.) We can also, for example, set $c = st$ so that we want $r$ to divide $(st)^{n-4}(s^n + t^n)$ which produces many other solutions, e.g. setting $r = s^n + t^n$ gives a $2$-parameter family of solutions
$$(x, y, z) = st(s^n + t^n, s, t), s, t, s^n + t^n \in \mathbb{Z} \setminus \{ 0 \}$$
and similarly for $c = rs, c = rt$. We could also set $c = r, s, t$ and then we'd get the two-variable version of this problem which should be pretty straightforward too.
For $n = 3$ scaling is a symmetry of the problem so it's natural here to ask for solutions such that $\gcd(x, y, z) = 1$. For $n = 2$ we expect that for large values of $x, y, z$ all about the same size the product $xyz$ is going to be larger than $x^2 + y^2 + z^2$ so at least one of $x, y, z$ has to be somewhat small compared to the other two for solutions to exist.
