How to prove $n=7k$ or $7k+1$, if $n=x^2=y^3$. $n$ is both a square and a cube, I need to prove $n=7k$ or $n=7k+1$, for some integer $k$.
I need some guide lines.

If $n$ is both a square and a cube, it is a perfect $6$-th power. 
Let $z$ be a number. Then $z$ has shape $7k$, or $7k+1$, or $7k+2$, and so on up to $7k+6$.
Then $z^6$ has shape (respectively) $7k$, $7k+1$, $7k+1$, $7k+1$, $7k+1$, $7k+1$, and $7k+1$.
To do the calculations, let us take the example $z$ of the shape $7k+2$.
Imagine expanding $(7k+2)^6$ using the Binomial Theorem. The first $6$ terms are obviously divisible by $7$, and the last term is $2^6=64$. This has remainder $1$ on division by $7$. So $(7k+2)^6$ has remainder $1$ on division by $7$, and therefore has shape $7t+1$.
The other calculations are very similar.
Remark: We do not need to work with $6$-th powers. The squares of numbers of shape $7k+a$, where $a$ goes from $0$ to $6$, are, in order, of shape
$$7k \qquad 7k+1 \qquad 7k+4 \qquad 7k+2\qquad 7k+2 \qquad 7k+4\qquad 7k+1.$$
The cubes, are, in order,  of the shape
$$7k \qquad 7k+1 \qquad 7k+1 \qquad 7k+6\qquad 7k+1 \qquad 7k+6\qquad 7k+6.$$
Only $7k$ and $7k+1$ occur in both lists. 
