A good place to start is to set [tex] \sqrt{x} [/tex] to y. That would mean we are looking for [tex] \sqrt{120-y} [/tex] to be an integer. Clearly, [tex] y\leq 120 [/tex], because if y were greater the part under the radical would be a negative, making the radical an imaginary number, not an integer. Also note that since [tex] \sqrt{x} [/tex] is a radical, it only outputs values from [tex] [0,\infty] [/tex], which means y is on the closed interval: [tex] [0,120] [/tex].
With that, we don't really have to consider y anymore, since we know the interval that [tex] \sqrt{x} [/tex] is on.
Now, we don't even have to find the x values. Note that only 11 perfect squares lie on the interval [tex] [0,120] [/tex], which means there are at most 11 numbers that x can be which make the radical an integer. All of the perfect squares are easily constructed. We can say that if k is an arbitrary integer between 0 and 11 then:
[tex] \sqrt{120-\sqrt{x}}=k \implies \\
\sqrt{x}=k^2-120 \implies\\
x=(k^2-120)^2 [/tex]
Which is strictly positive so we know for sure that all 11 numbers on the closed interval will yield a valid x that makes the radical an integer.
