I thought it would be instructive to present a way forward that relie on only pre-calculus analysis including some elementary inequalities. To that end we now proceed.
PRIMER:
In THIS ANSWER, I showed using elementary, pre-calculus tools that the arctangent function satisfies the inequalities
$$\frac{x}{\sqrt{1+x^2}}\le \arctan(x)\le x\tag1$$
for $x\ge 0$. We now use $(1)$ in the development that follows.
Enforcing the substitution $x\mapsto 1/x$ reveals
$$\begin{align}
\lim_{x\to \infty}\left(\frac2\pi \arctan(x)\right)^{\frac{x^2}{1+2x}}&\overbrace{=}^{x\mapsto 1/x}\lim_{x\to 0}\left(\frac2\pi \arctan(1/x)\right)^{\frac1{x(x+2)}}\\\\
&=\lim_{x\to 0}\left(1-\frac2\pi \arctan(x)\right)^{\frac1{x(x+2)}}\tag2
\end{align}$$
Using $(1)$ along with the inequality $\frac1{\sqrt{1+x^2}}\ge 1-\frac12 x^2$ in $(2)$, we find that
$$\left(1-\frac2\pi \left(x-\frac12x^3\right)\right)^{\frac1{x(x+2)}}\le \left(1-\frac2\pi \arctan(x)\right)^{\frac1{x(x+2)}}\le \left(1-\frac2\pi x\right)^{\frac1{x(x+2)}}\tag3$$
Next, recalling that $\lim_{x\to 0}\left(1+ tx\right)^\frac1x=e^t$, it is easy to see that $\lim_{x\to 0}\left(1-\frac2\pi x\right)^{\frac1{x(x+2)}}=e^{-1/\pi}$.
We will now show that the limit of the left-hand side of $(3)$ is also $e^{-1/\pi}$, whence application of the squeeze theorem yields the coveted limit.
Proceeding, we write the left-hand side of $(3)$ as
$$\begin{align}
\left(1-\frac2\pi \left(x-\frac12x^3\right)\right)^{\frac1{x(x+2)}}&=\left(1-\frac2\pi x\right)^{\frac1{x(x+2)}}\times \color{blue}{\left(1+x\,\frac{x^2}{\pi-2 x}\right)^{\frac1{x(x+2)}}}\tag4
\end{align}$$
Using Bernoulli's Inequality, we have for $0<x<\pi$
$$\begin{align}
1\le \color{blue}{\left(1+x\,\frac{x^2}{\pi-2 x}\right)^{\frac1{x(x+2)}}}&\le \frac1{\left(1-x\,\frac{x^2}{\pi-2 x}\right)^{\frac1{x(x+2)}}}\\\\
&\le \frac1{1-\frac{x^2}{(x+2)(\pi -2x)}}\tag5
\end{align}$$
Applying the squeeze theorem to $(5)$, we find that
$$\lim_{x\to 0}\left(1+x\,\frac{x^2}{\pi-2 x}\right)^{\frac1{x(x+2)}}=1\tag6$$
Using $(6)$ in $(4)$ reveals
$$\lim_{x\to 0}\left(1-\frac2\pi \left(x-\frac12x^3\right)\right)^{\frac1{x(x+2)}}=e^{-1/\pi}\tag7$$
Finally, using $(7)$ in $(3)$ and then equating to $(2)$ yields the coveted limit
$$\lim_{x\to \infty}\left(\frac2\pi \arctan(x)\right)^{\frac{x^2}{1+2x}}=e^{-1/\pi}$$
And we are done!
Tools Used: Elementary pre-calculus analysis (e.g. Bernoulli's Inequality and other elementary inequalities) only along with the limit definition of the exponential function.