Most (if not all) solutions focus on this as a problem for these specific numbers. And, in fact, the same question was then asked for other specific pairs of numbers. Moreover, most of the solutions seemed to be quite complicated. So, I thought I would post my solution here.

First, since the natural log is an increasing function, we have for any ##a, b > 0##:
$$a^b > b^a \Leftrightarrow b\ln(a) > a\ln(b) \Leftrightarrow \frac{\ln(a)}a > \frac{\ln(b)}b$$We look at the function ##f(x) = \frac{\ln(x)}x## and note that:
$$f'(x) = \frac 1 {x^2}(1 - \ln(x))$$We see that the function is increasing for ##x < e##, has a maximum at ##x=e##, and is decreasing for ##x > e##. In particular, we can see that:

If ##e \le a < b##, then ##a^b > b^a##. (That answers the question for ##e## and ##\pi##)

If ##a < b \le e##, then ##a^b < b^a##.

My recollection was the famous exact solution of ##2^4 = 4^2##

I guess he wanted to find algebraically the point where the a and b were equal inflection point, which I think is ##e^e##

My high school friend in the 1960s played with this problem for a long time. I never understood what he was trying to determine. He was an MAA champion and was on the team that went to England to compete with the English and Russians. The US team came in last because they were used to multiple-choice tests, while the English and Russians were trained on fill-in-the-blank tests, which is how the competition test was constructed.

My recollection was the famous exact solution of ##2^4 = 4^2##

I guess he wanted to find algebraically the point where the a and b were equal inflection point, which I think is ##e^e##

The remaining case to investigate from OP is a<e<b.

Some cases
$$2^2=2^2$$
$$2^3<3^2$$
$$2^4=4^2$$
$$2^5>5^2$$

We can get two roots of the equation
$$\frac{\ln x}{x}=c < \frac{1}{e}$$
$$e^{cx}=x$$
by the graphs.

[EDIT]
According to the wikipedia article : https://en.wikipedia.org/wiki/Lambert_W_function#Generalizations the solutions of
$$x=e^{cx}$$
is
$$x=-\frac{1}{c}W(-c)=x_1(c),x_2(c)$$
where W is Lambert function of branch 0 and -1. Thus with parameter 0<c<1/e
$$x_1^{x_2}=x_2^{x_1}$$

As a layman I should appreciate any suggestion/correction.

When you think of it, there are endless possible definitions for such differences or additions; and I guess you can also use these definitions in theoretical physics.
But nice analasys for the case of numbers.

As for a similar case for matrices, I guess one needs an inner product to define positive definiteness, and then we can do a similar analysis for ##A^B## and ##B^A##; I would think Hessians will appear.. (haven't done the analysis myself it's just a hunch). :

Though, it appears to develop brain flatulence if the numbers get too large.

Apple's 'grapher' application also generates it.

View attachment 369812

Though, it appears to develop brain flatulence if the numbers get too large.

View attachment 369813

Also, it’s obvious but if you look at the regions where a^b >b^a vs <, they are symmetric about the a=b axis so exactly half the quadrant is greater and half is lesser.