Find a non-negative integer $ x $ such as $ sqrt {x ^ 2 + sqrt {x + 1}} $ is a positive integer

Because $ sqrt {x ^ 2 + sqrt {x + 1}}> x $, we leave $ x ^ 2 + sqrt {x + 1} = (x + y) ^ 2, (y> 0) $

That means $$ begin {aligned}

& x ^ 2 + sqrt {x + 1} = x ^ 2 + y ^ 2 + 2xy \

& implies sqrt {x + 1} = y ^ 2 + 2xy \

& implies x + 1 = y ^ 4 + 4x ^ 2y ^ 2 + 4xy ^ 3

end {aligned} $$

And that 's where I was stuck.