Unlike the famous question of comparing $e^\pi$ and $\pi^e$, which I solved almost instantly, I am stuck with this problem. My thought was the following.
Since exponential function is order-preserving, we exponentiate both terms and get $\pi$ and $e^{\pi-2}$. Then we study the function $f(x) = e^{x-2} - x$ or the function $g(x) = \frac{e^{x-2}}{x}$, and compare them with zero and one respectively. I tried both. But both involve solving the equation $$e^{x-2} = x.$$
I tried Lagrange error terms and have $$f(x) = -1 + \frac{(x-2)^2}{2!} + R_2(x-2),$$ where $$\frac{(x-2)^3}{3!} \le R_2(x-2) \le \frac{e^{x-2}}{3!} (x-2)^3.$$
It is easy to see that the equation have a root between $3$ and $2 + \sqrt2$. But I don't know how close it is to $\pi$. It is to provide some lower bounds since we can plug in some values and calculate to show that $f(x) > 0$ for such values. But for the upper bound, it is hard to calculate by hands since it has the $e^{x-2}$ factor. At my best attempt by hand, I showed that $f(3.15) > 0$. All it entails is that for all $x \ge 3.15$, $e^{x-2}$ is greater than $x$. But it tells nothing about the other side.
Then I looked at the calculator and find that $e^{\pi-2} < \pi$.
I also tried Newton-Raphson iteration, but it involves a lot of exponentiation which is hard to calculate by hand and also involves approximation by themselves. And I don't know how fast and close the iteration converges to the true root of the equation.
Any other hint for comparing these two number purely by hand?