That $m$ Under The Radicals

$(1+\sqrt{m})^n = 2(1+\sqrt{m})^{n-1} + (1+\sqrt{m})^{n-2}\\ (1+\sqrt{m})^n - 2(1+\sqrt{m})^{n-1} - (1+\sqrt{m})^{n-2} = 0\\ (1+\sqrt{m})^{n-2} \big((1+\sqrt{m})^2 - 2(1+\sqrt{m}) -1\big)=0$

Now, note that for any positive integer $m, \;1+\sqrt{m} > 1$

This means that $(1+\sqrt{m})^{n-2} >0$ for any real value of $n$

Since this factor cannot be $0$ , it means that the other factor $(1+\sqrt{m})^2 - 2(1+\sqrt{m}) -1$ must be $0$

$(1+\sqrt{m})^2 - 2(1+\sqrt{m}) -1=0\\ 1+2\sqrt{m}+m -2 - 2\sqrt{m}-1=0\\ m-2=0\\ m=\boxed{2}$

Since the identity has to be true for any $n$ value, we can choose $n = 2$ . The identity becomes: $\left(1+\sqrt{m}\right)^2=2\left(1+\sqrt{m}\right)^{1}+\left(1+\sqrt{m}\right)^{0}$ $= \not{1} + m + \cancel{2\sqrt{m}} = 2 + \cancel{2\sqrt{m}} + \not{1}$ $\Rightarrow m = \boxed{2}$