Finding $e$ through $a$

$\begin{aligned} \lim_{n \rightarrow \infty} \left (\dfrac{n + a}{n - a} \right)^n &= e \\ \ln \left ( \lim_{n \rightarrow \infty} \left (\dfrac{n + a}{n - a} \right)^n \right ) &= \ln e \\ \lim_{n \rightarrow \infty} \ln \left (\dfrac{n + a}{n - a} \right)^n &= 1 \\ \lim_{n \rightarrow \infty} n \ln \dfrac{n + a}{n - a} &= 1 \\ \lim_{n \rightarrow \infty} \dfrac{\ln \frac{n +a}{n - a}}{\frac{1}{n}} &= 1. \end{aligned}$

The left hand side limit is of the form $\dfrac{0}{0},$ so we can apply L'Hopital's rule:

$\begin{aligned} \lim_{n \rightarrow \infty} \dfrac{\ln (n+a) - \ln (n - a)}{\frac{1}{n}} &= 1 \\ \lim_{n \rightarrow \infty} \dfrac{\frac{1}{n + a} - \frac{1}{n - a}}{-\frac{1}{n^2}} &= 1 \\ \lim_{n \rightarrow \infty} \dfrac{-\frac{2a}{n^2 - a^2}}{-\frac{1}{n^2}} &= 1 \\ \lim_{n \rightarrow \infty} \dfrac{2an^2}{n^2 - a^2} &= 1 \\ 2a &= 1 \\ a &= \dfrac{1}{2}. \end{aligned}$

Thus, $m = 1, n = 2,$ and $\left (\sqrt{n + \sqrt{m}} \right)^n = \left (\sqrt{2 + \sqrt{1}} \right)^2 = \boxed{3}.$

$\begin{aligned} e & =\lim_{n \to \infty} \left(\frac {n+a}{n-a} \right)^n \\ & = \lim_{n \to \infty} \left(\frac {n-a+2a}{n-a} \right)^n \\ & = \lim_{n \to \infty} \left(1 + \frac {2a}{n-a} \right)^n \\ & = \lim_{n \to \infty} \left(1 + \frac 1{\color{#3D99F6}\frac {n-a}{2a}} \right)^{{\color{#3D99F6}\frac {n-a}{2a}} \times 2a + a} & \small \color{#3D99F6} \text{Let }u = \frac {n-a}{2a} \\ & = \lim_{\color{#3D99F6}u \to \infty} \left(1 + \frac 1{\color{#3D99F6}u} \right)^{2a{\color{#3D99F6}u}+ a} \\ \implies e & = e^{2a} \\ \implies a & = \frac 12 \end{aligned}$

$\implies \left(\sqrt{n + \sqrt m}\right)^n = \left(\sqrt{2 + \sqrt 1}\right)^2 = \boxed{3}$

Note: This is an invalid approach. See comments below

The following arithmetic holds when $n$ is very large (tending toward infinity)

$\displaystyle \lim_{n \to \infty} \left(\frac{n + a}{n - a}\right)^n = e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n \\ \frac{n + a}{n - a} = 1 + \frac{1}{n} \\ n + a = n - a + 1 - \frac{a}{n} \\ \frac{a}{n} \approx 0 \implies 2a \approx 1 \\ a \approx \frac{1}{2}$