You should test if the function `f(x) = x^3 - 3x` strictly increases over `(1,oo),` hence, you need to use the first derivative to check it out, such that:
`f'(x) = (x^3 - 3x)' => f'(x) = 3x^2 - 2`
Since `x in (1,oo)` yields that `3x^2 - 2 > 0 => f'(x) > 0` . Since `f'(x)>0` , then the function `f(x)` strictly increases over `(1,oo)` .
You need to evaluate `f(x)` at `x = 1` , such that:
`f(1) = 1^3 - 3*1 = -2`
Since `f(1) = -2 < 0` and the function strictly increases for `x > 1` , hence, the graph of the function intersects x axis only one time, thus, the equation `x^3 - 3x = m > -2` has only one solution over the interval `(1,oo)` , as the sketch below ilustrates it.