Modular arithmetic and using in well-ordering principle I need to prove the following, but I do not know how to go about it.
If $$   (*)\:\:\: x^{3} - y^{3}= 3^{n} $$
Then $$ x \equiv 0 (mod 3)  \:\: and \:\:\: y \equiv 0 (mod 3)$$
In addition, to prove that the equation above (*), there is no solution in $ N^{+} $ (in other words $ \quad 0\neq x,y\in \mathbb N $ ).
By using the Well-ordering principle, i.e. Any non-empty subset of N, the set of natural numbers has a minimum element.
Edit
I can use these facts: $$ \forall x\in \mathbb N^{+}  \:\:\: x^{3}  \equiv x (mod 3)$$
And also:
$$ \forall x,y\in \mathbb N^{+} , \:\:\: if \:\:\: x^{3} - y^{3}\equiv 0 (mod 3) \:\:\: \\Then \:\:\: x \equiv y (mod 3)  $$
Thanks
 A: Let $x$ and $y$ be positive integers such that $x^3-y^3=3^n$, where $n\ge 3$. We first show  that $3$ divides $x$ and $3$ divides $y$. Then later we use this to show that in fact there are no positive solutions. 
In the OP, it is observed that $x\equiv y\pmod{3}$, so $x-y$ is divisible by $3$.  Now we use the fact that $x^3-y^3=(x-y)(x^2+xy+y^2)$. 
Note that if $x$ and $y$ are positive and not both equal to $1$, it follows that $x^2+xy+y^2\gt 3$, and therefore $x^2+xy+y^2=3^k$ for some $k\ge 2$. Now from the fact that $9$ divides $(x-y)^2$ and the identity 
$$(x^2+xy+y^2)-(x-y)^2=3xy$$
we conclude that $9$ divides $3xy$.  Thus $3$ divides $xy$. It follows that $3$ divides one of $x$ or $y$, and therefore since $x\equiv y\pmod{3}$, it divides both.

Now that we have the divisibility result, the rest follows by the least number principle, or equivalently by infinite descent.
It is easy to verify that there are no positive solutions with $n=0$, $n=1$, or $n=2$. Suppose that for some $n\ge 3$ there are positive solutions. Then there is a smallest $n\ge 3$ with this property. Let $x^3-y^3=3^n$. From our divisibility result, we have $x=3s$, $y=3t$ for some positive $s$ and $t$. But then $3^3s^3-3^3t^3=3^n$ and therefore $s^3-t^3=3^{n-3}$, contradicting the minimality of $n$.
Remark: It is not quite true that if $x$ and $y$ are integers such that $x^3-y^3$ is a power of $3$ then $3$ divides $x$ and $y$. There is the obvious example $x=1$, $y=0$. But there is also the more interesting example $x=2$, $y=-1$, for which we have $x^3-y^3=9$.
