Prove that if an integer is a perfect square and a perfect cube, simultaneously, then it is either of the form 7k or 7k+1

If an integer is both a square and a cube, it can be of the form:
(a3)^2
Now,
since a cube can be of the form 7k or 7k+-1(thanks to FoolForMath),
 we write
a^3=7k
and get the no to be
49k^2
, which is in the form of 7 times something
49k^2=7×(7k^2)
Now put
a^3=7k+−1 Square it
and you'll get a number in the form of (7times something +1)