Solution by radicals of $(1+x)^n=x^m$
The difference of powers formula can yield solutions for $m>4$: $0 = (1+x)^{n} - x^{m} = (1+x)^{n} - (x^\frac{m}{n})^{n} = (1+x-x^\frac{m}{n})((1+x)^{n-1} +(1+x)^{n-2}x^{\frac{m}{n}} + \ldots + (x^{\frac{m}{n}})^{n-1})$.
Under the assumption that $m>n$, then the zeros $1+x-x^{\frac{m}{n}}$ will yield all possible solutions. The huge benefit here is that zeros of this polynomial can be quickly found using Newton's Method. This can help you find your desired other solutions.