Here, $^yx$ means "$x$ tetrated to $y$", and is defined recursively by
$^yx = \begin{cases} 1, & \text{if y = 0}\\ x^{^{y-1}x}, & \text{if y > 0} \end{cases}$
I want to know if there are non-trivial positive integer solutions to $^ba$ = $^dc$. I define a solution to be trivial if $\min\{a, b, c, d\} = 1$ or $a =c$.
I have already proven that there are no solutions where $\{b, d\} = \{2, 3\}$, and it is as follows:
Suppose that $x^{x^x}=y^y$ for positive integers $x, y > 1$. Then $x$ and $y$ are both perfect powers of a common integer, so let $x = p^a$, $y = p^b$ (note that $a < b$). Then we get
$p^{ap^{ap^a}}=p^{bp^b}$, and therefore, $ap^{ap^a}=bp^b$, or $\frac{b}{a}=\frac{p^{ap^a}}{p^b}$
So let $b = ap^c$ for some positive integer $c$. Our equation turns to $p^{ap^a} = p^{b+c}$, or $ap^a = ap^c + c$ (note this means $a > c$). Since all variables are positive integers and $p > 1$, we know that $ap^c > c$, so $ap^c + c < 2ap^c \le ap^{c+1} \le ap^a$, but we determined that $ap^c + c = ap^a$, so we have a contradiction.
I have not made any progress towards solving the general equal tetration case beyond here. If someone can prove that there are no solutions for larger $b$ and $d$ (which is what I suspect), this would prove that addition, multiplication, and exponentiation are the only hyper-operations $*$ where $a*b = c*d$ has non-trivial solutions among the positive integers.
There are no non-trivial solutions to ${^b{a}}={^d{c}}$, as shown in my first-ever attempt at a maths paper :-) https://www.preprints.org/manuscript/201902.0176/v1
The proof is a little too long to recreate here, I'm afraid.