Three, (50 points) Let $s$ be a natural number greater than 1. Find all natural numbers $n$ such that for $n$ there exist coprime natural numbers $x$ and $y$ satisfying $3^{n}=x^{s}+y^{s}$.
Three, first prove a lemma.
Lemma Let $x, y, p, n, k \in \mathbf{N}$, and satisfy $x^{n}+y^{n}=p^{k}$. If $n$ is an odd number greater than 1, and $p$ is an odd prime, then $n$ can be expressed as a power of $p$ with a natural number as the exponent.

Proof of the lemma: Let $m$ be the greatest common divisor of $x, y$, and we can set $x = m x_{1}, y = m y_{1}$. From the given condition, we have
$$
m^{n}\left(x_{1}^{n}+y_{1}^{n}\right)=p^{k}.
$$

Therefore, there exists a non-negative integer $\alpha$ such that
$$
x_{1}^{n}+y_{1}^{n}=p^{k-n}.
$$

Since $n$ is odd, we have
$$
\frac{x_{1}^{n}+y_{1}^{n}}{x_{1}+y_{1}}=x_{1}^{n-1}-x_{1}^{n-2} y_{1}+x_{1}^{n-3} y_{1}^{2}-\cdots-x_{1} y_{1}^{n-2}+y_{1}^{n-1}.
$$

Let $A$ denote the number on the right side of the equation. Since $p > 2$, at least one of $x_{1}$ and $y_{1}$ is greater than 1. And since $n > 1$, we have $A > 1$. From equation (1), we get
$$
A\left(x_{1}+y_{1}\right)=p^{k-n}.
$$

Since $x_{1}+y_{1} > 1$ and $A > 1$, they can both be divided by $p$, and there exists a natural number $\beta$ such that $x_{1}+y_{1}=p^{\beta}$. Thus,
$$
\begin{aligned}
A= & x_{1}^{n-1}-x_{1}^{n-2}\left(p^{\beta}-x_{1}\right)+x_{1}^{n-3}\left(p^{\beta}-x_{1}\right)^{2}- \\
& \cdots-x_{1}\left(p^{\beta}-x_{1}\right)^{n-2}+\left(p^{\beta}-x_{1}\right)^{n-1}
\end{aligned}
$$
$=n x_{1}^{n-1}+B p$ (where $B$ is some integer).
Since $p \mid A$ and $\left(x_{1}, p\right)=1$, we have $p \mid n$.
Let $n=p q$, then $x^{p q}+y^{p q}=p^{k}$, i.e.,
$$
\left(x^{p}\right)^{q}+\left(y^{p}\right)^{q}=p^{k}.
$$

If $q>1$, by the same proof as above, we can show that $q$ can be divided by $p$. If $q=1$, then $n=p$; repeating this process, we can conclude that there exists a natural number $l$ such that $n=p^{l}$.
Now we prove the conclusion of the problem: the possible value of $n$ is only 2.
Let $3^{n}=x^{3}+y^{3}$, where $(x, y)=1$, and assume $x>y$. Since $s>1, n \in \mathbf{N}$, it is clear that $3 \nmid x$ and $3 \nmid y$. We discuss as follows:
(1) If $s$ is even, then
$$
x^{3} \equiv 1(\bmod 3), y^{3} \equiv 1(\bmod 3).
$$

Thus, $x^{3}+y^{3} \equiv 2(\bmod 3)$, which means $x^{3}+y^{3}$ is not an integer power of 3, a contradiction.
(2) If $s$ is odd and $s>1$, then
$$
3^{n}=(x+y)\left(x^{s-1}-x^{s-2} y+\cdots+y^{s-1}\right).
$$

Thus, $x+y=3^{m}, m \geqslant 1$. We now prove that $n \geqslant 2 m$.
By the lemma, $3 \mid s$. Taking $x_{1}=x^{\frac{3}{3}}, y_{1}=y^{\frac{1}{3}}$, we can assume $s=3$. Thus, $x^{3}+y^{3}=3^{n}, x+y=3^{m}$.
Therefore, we only need to prove that $x^{3}+y^{3} \geqslant(x+y)^{2}$, i.e., prove that $x^{2}-x y+y^{2} \geqslant x+y$.
Since $x \geqslant y+1$, we have $x^{2}-x=x(x-1) \geqslant x y$.
Thus, $\left(x^{2}-x y-x\right)+\left(y^{2}-y\right) \geqslant 0$.
Therefore, $n \geqslant 2 m$ is proved.
From $(x+y)^{3}-\left(x^{3}+y^{3}\right)=3 x y(x+y)$, we get $3^{2 m-1}-3^{n-m-1}=x y$.
And $2 m-1 \geqslant 1$ and $n-m-1 \geqslant n-2 m \geqslant 0$.
If at least one of the inequalities in (2) is strict, then
$n-m-1=n-2 m=0$.
Then, $m=1, n=2$ and $3^{2}=2^{3}+1^{3}$.
Thus, $n=2$ is the only value that satisfies the condition.