I have not thought about this particular problem yet, but let me prove somewhat related claim that if $x,y$ are integers and $x^n-1$ divides $y^n-1,$ $y>1$ for all $n\in\mathbb{N}$ (this condition can be eventually relaxed) then $y=x^m$ for some $m\in\mathbb{N}.$ This, in particular implies that both $x$ and $y$ cannot be primes simultaneously.
The key is the following useful lemma.
Lemma. If $x_1....x_m\ge 1$ are distinct rational numbers and $\alpha_1,...\alpha_m\ne 0$ are real numbers such that $\lim_{n\to\infty}\sum_{k=1}^m\alpha_kx_k^n-m_n=0$ where $m_n\in\mathbb{N},$ then $x_i\in\mathbb{Z},$ for $1\le i\le m.$
Proof. Induction on $m.$ The base case is fairly interesting exercise, I will leave it for the reader. To prove a step of induction from $m$ to $m+1,$ we fix some $x_l=\frac{p}{q}.$ Observe, that
$$\lim_{n\to\infty}\sum_{k=1}^mp\alpha_kx_k^n-pm_n=0$$
and
$$\lim_{n\to\infty}\sum_{k=1}^mq\alpha_kx_k^{n+1}-qm_{n+1}=0.$$
Subtracting last two identities leads to
$$\lim_{n\to\infty}\sum_{k=1}^m\alpha_kx_k^n(qx_k-p)-(qm_{n+1}-pm_n)=0.$$
Since $x_lq-p=0,$ we can apply induction hypothesis to conclude that $x_i\in\mathbb{N}$ for $i\ne l.$ We are left to note that $l$ was chosen arbitrary and thus, $x_i\in\mathbb{N}$ for $1\le m+1.$
Now, in order to apply this to our problem we pick such $m$ that $x^m\le y<x^{m+1}$ and take $m_i=\frac{y^i-1}{x^i-1},$ $x_i=\frac{y}{x^i},$ $\alpha_i=1$ for $1\le i\le m.$ Then
$$m_n-\sum_{k=1}^m\alpha_kx_k^n=\frac{y^n-x^{mn}}{x^{mn}(x^n-1)}\to 0.$$
So all $x_i$ are integers. Thus, $$\frac{y^n-x^{mn}}{x^{mn}(x^n-1)}=0$$
for sufficiently large $n.$ Therefore $y=x^m.$
Note, that we can request $\lim_{n\to\infty}\sum_{k=1}^m\alpha_kx_k^n-m_n=0$ along some "fairly nice" subsequence $n_k.$ Thus, we do not require divisibility $x^n-1$ and $y^n-1$ for all $n.$
Remark 1. As to the second question, I do not think that there is something special about $n,$ in the sense that it can be fairly large. Good computer search should help to find a counterexample.
Remark 2. Starting from any prime $q$ and any $a>1,$ and $q^n-1$ divides $a^n-1$ we can take $x_m=a+m(q^{n}-1)$ and apply Dirichlet's theorem for arithmetic progressions to find infinitely many primes that satisfy our condition.
First, observe that we should have $x\le12$. Otherwise:
$$x^3(y^3+z^3)\ge x^6+x^3(xyz)\ge 13^6+2197(xyz)\gt2012(xyz+2).$$
Therefor $x \in \{1,2,3,4,5,6,7,8,9,10,11,12\}$.
But $x$ cannot be $3, 5,6 , 7, 9, 10, 11$ or $12$ because:
$$x|2012(xyz+2) \implies x|2012 \times2 \implies x|8 \times 503.$$
On the other hand $x$ cannot be $4$ or $8$ either because, in this case, we should have:
$$4^3|4 \times 503(4yz+2) \ or \ {} 8^3|4 \times 503(8yz+2),$$
both of which are impossible.
Therefore $x$ is either $1$ or $2$, and we have two new equations:
$$y^3+z^3=2012(yz+2) \\ y^3+z^3=503(yz+1).$$
Now, note that if $503|y$ then we get $503|z$, hence $503|y+z.$
Let's assume $503 \not| \ y$. Hence $503 \not| \ z$. By the Fermat's theorem, we have:
$$503|z^{502} - y^{502};$$
and because of $503| y^3+z^3$ we conclude that $503| y^{501}+z^{501}$, and as a result $503|y^{502}+yz^{501}$. Therefore:
$$503|z^{502}+yz^{501}\implies 503|y+z \implies y+z=503k.$$
Now, we have two other new equations:
$$ (1) \ k(y-z)^2+(k-4)yz=8 \\ (2)\ k(y-z)^2+(k-1)yz=1.$$
In the first case, it is very easy to see that the equation has no solution. First notice that $k$ should be $2$, and then we have $(y+z)^2-5yz= 4$ while $y+z=503 \times 2$. This means that the equation has no solution.
For the second equation, it is clear that $k$ must be equal to either $1$ or $2$. If $k=2$ then $z=y=1$, which is impossible. Thus $k=1$, and the only solution is $(2,251,252).$
EDIT: This problem is a shortlist problem of IMO $2012$ and has an official solution.
Best Answer
When $n$ is even, any odd prime $p$ dividing $x+y$ does not divide $x^n+y^n=x^n-y^n+2y^n$ because $p$ divides $x^n-y^n$ and $y$ and $p$ are coprime.
So remaining case is $y=2^k-x$ where $x$ is odd. When $n$ is odd, you can use LTE. When $n$ is even, you can show that $x^n+y^n$ can be divided by $2$ only once because $x,y$ are odd.