Write the integers as $x$ and $x + 60.$ If $\sqrt{x} + \sqrt{x + 60} = \sqrt{y}$ with $y$ an integer, then $x(x + 60)$ must be a perfect square

Set $x(x + 60) = z^2$ and factor $(x + 30 + z)(x + 30 - z) = 900;$ discard the case where both integers are perfect squares

Let the integers be $x$ and $x + 60,$ and suppose $\sqrt{x} + \sqrt{x + 60} = \sqrt{y}.$ Squaring, $y = 2x + 60 + 2\sqrt{x(x + 60)},$ so $x(x + 60)$ must be a perfect square, say $z^2.$ Completing the square, $$(x + 30)^2 - z^2 = 900,$$ i.e. $$(x + 30 + z)(x + 30 - z) = 900.$$ The two factors have the same parity and their product is even, so both are even.

The factor pairs $(450, 2),$ $(150, 6),$ $(90, 10),$ $(50, 18)$ give $x + 30 = 226,$ $78,$ $50,$ $34,$ so $x = 196,$ $48,$ $20,$ $4.$ For $x = 196$ the integers are $196$ and $256,$ both perfect squares, so $\sqrt{y} = 14 + 16 = 30$ and $y = 900$ is a perfect square — not allowed. For $x = 48$ the integers are $48$ and $108,$ with $\sqrt{48} + \sqrt{108} = 4\sqrt{3} + 6\sqrt{3}$ $= \sqrt{300},$ and $300$ is not a perfect square.

The maximum possible sum is therefore $48 + 108 = 156.$