$n$ is a square and a cube $a^2 = n = b^3\Rightarrow n\equiv 0,1\pmod{7}$ 
Verify that if an integer is simultaneously a square and a cube, then it must be either of the form ${7k}$ or ${7k +1}$.

I have no idea on how to proceed.

If an integer $a$ is simultaneously a square and a cube, it must be the sixth power of an integer $b$
Now as $7$ is prime either $7|b$ or $(7,b)=1$
If $7|b, 7|b^n$ for integer $n\ge1$
Else by Fermat's Little Theorem , $7|(b^{7-1}-1)$
