I like the compact solution you found, but:
Since you use the "[i ** 2 for i in array1]" part you first generate a full list in memory, then check if it is the same as the other list.
For long lists it is far more efficient to generate the squares element by element while checking, because you could find a different early on and just return false.

Hi man, one hint:
You can get rid of solution a1 because that is always >1, therefore it is not a converged solution.