Find all positive natural $x$ and $y$ so $(x+y)^7-x^7-y^7$ is divisble by $7^7$, but so that $xy(x+y)$ is not divisble by $7$ I thought of looking modulo $7$ but it lead me nowhere... Does anyone have a better idea?

Note that 
$$\frac{(x+y)^7-x^7-y^7}{7xy(x+y)}= x^4 + 2yx^3 + 3y^2x^2 + 2y^3x + y^4 $$
We immediately check that $x\equiv y\pmod 7$ cannot lead to a solution because $9x^4$ cnnot be a multiple of $7^6$. Hence we are allowed to rewrite this as
$$x^4 + 2yx^3 + 3y^2x^2 + 2y^3x + y^4 =\frac{x^6-2x^3y^3+y^6}{(x-y)^2}=\frac{(x^3-y^3)^2}{(x-y)^2}. $$
In order to have $7^6\mid (x^3-y^3)^2$, we need $7^3\mid x^3-y^3=(x-y)(x^2+xy+y^2)$ and by the bove remark $7^3\mid x^2+xy+y^2$.
As $7\nmid x,y$, there exists $z$ with $y\equiv zx\pmod {7^3}$. Then we need to solve $z^2+z+1\equiv 0\pmod{7^3}$.
Modulo $7$, we find that $z\equiv 2\lor 4\pmod 7$. By symmetry $x\leftrightarrow y$, we need only investigate the first case. Substituting $z=7w+2$, we arrive at
$$ 7w^2+5w+1\equiv 0\pmod {7^2}$$
In particular, $w\equiv 4\pmod 7$. With $w\leftarrow 7v+4$,
$$ 49v^2+61v+19\equiv  0\pmod 7$$
which simplifies to $v\equiv -1\pmod 7$.
Thus $v=7u-1$, $w=49u-3$, $z=343u-19$
