The Catalan numbers $C_n$ count both
 the Dyck paths of length $2n$, and
 the ways to associate $n$ repeated applications of a binary operation.
We call the latter magma expressions; we will explain below.
Dyck paths, and their lattice structure
A Dyck path of length $2n$ is a sequence of $n$ upandright strokes and $n$ downandright strokes, all having equal length, such that the sequence begins and ends on the same horizontal line and never passes below it. A picture of the five length6 Dyck paths is shown here:
A: B: C: D: E:
/\
/ \ /\/\ /\ /\
/ \ / \ / \/\ /\/ \ /\/\/\
There is an order relation on the set of length$2n$ Dyck paths: $P\leq Q$ if $P$ fits completely under $Q$; I'll call it the height order, though in the title of the post, I called it "Dyck order". I've been told it should be called the Stanley lattice order. For $n=3$ it gives the following lattice:
A

B
/ \
C D
\ /
E
For any $n$, one obtains a poset structure on the set of length$2n$ Dyck paths using height order, and in fact this poset is always a Heyting algebra (it represents the subobject classifier for the topos of presheaves on the twisted arrow category of $\mathbb{N}$, the free monoid on one generator; see this mathoverflow question).
Magma expressions and the "exponential evaluation order"
A set with a binary operation, say •, is called a magma. By a magma expression of length $n$, we mean a way to associate $n$ repeated applications of the operation. Here are the five magma expressions of length 3:
A: B: C: D: E:
a•(b•(c•d)) a•((b•c)•d) (a•b)•(c•d) (a•(b•c))•d ((a•b)•c)•d
It is wellknown that the set of length$n$ magma expressions has the same cardinality as the set of length$2n$ Dyck paths: they are representations of the $n$th Catalan number.
An ordered magma is a magma whose underlying set is equipped with a partial order, and whose operation preserves the order in both variables. Given an ordered magma $(A,$•$,\leq)$, and magma expressions $E(a_1,\ldots,a_n)$ and $F(a_1,\ldots,a_n)$, write $E\leq F$ if the inequality holds for every choice of $a_1,\ldots,a_n\in A$. Call this the evaluation order.
Let $P=\mathbb{N}_{\geq 2}$ be the set of natural numbers with cardinality at least 2, the logarithmically positive natural numbers. Equipped with the operation given by exponentiation, $c$•$d\:=c^d$, we obtain an ordered magma, using the usual $\leq$order. Indeed, if $2\leq a\leq b$ and $2\leq c\leq d$ then $a^c\leq b^d$.
Question: Is the exponential evaluation order on length$n$ expressions in the ordered magma $(P,$^$,\leq)$ isomorphic to the height order on length$2n$ Dyck paths?
I know of no a priori reason to think the answer to the above question should be affirmative. A categorical approach might be to think of the elements of $P$ as sets with two special elements, and use them to define injective functions between Homsets, e.g. a map $$\mathsf{Hom}(c,\mathsf{Hom}(b,a))\to\mathsf{Hom}(\mathsf{Hom}(c,b),a).$$ However, while I can define the above map, I'm not sure how to generalize it. And the converse, that being comparable in the exponential evaluation order means that one can define a single injective map between homsets, is not obvious to me at all.
However, despite the fact that I don't know where to look for a proof, I do have evidence to present in favor of an affirmative answer to the above question.
Evidence that the orders agree
It is easy to check that for $n=3$, these two orders do agree:
a^(b^(c^d)) A := A(a,b,c,d)
 
a^((b^c)^d) B
/ \ / \
(a^b)^(c^d) (a^(b^c))^d C D
\ / \ /
((a^b)^c)^d E
This can be seen by taking logs of each expression. (To see that C and D are incomparable: use a=b=c=2 and d=large to obtain C>D; and use a=b=d=2 and c=large to obtain D>C.) Thus the evaluation order on length3 expressions in $(P,$^$,\leq)$ agrees with the height order on length $6$ Dyck paths.
(Note that the answer to the question would be negative if we were to use $\mathbb{N}$ or $\mathbb{N}_{\geq 1}$ rather than $P=\mathbb{N}_{\geq2}$ as in the stated question. Indeed, with $a=c=d=2$ and $b=1$, we would have $A(a,b,c,d)=2\leq 16=E(a,b,c,d)$.)
It is even easier to see that the orders agree in the case of $n=0,1$, each of which has only one element, and the case of $n=2$, where the order $(a^b)^c\leq a^{(b^c)}$ nottoosurprisingly matches that of length4 Dyck paths:
/\
/\/\ ≤ / \
Indeed, the orderisomorphism for $n=2$ is not too surprising because there are only two possible partial orders on a set with two elements. However, according to the OEIS, there are 1338193159771 different partial orders on a set with $C_4=14$ elements. So it would certainly be surprising if the evaluation order for length4 expressions in $(P,$^$,\leq)$ were to match the height order for length8 Dyck paths. But after some tedious calculations, I have convinced myself that these two orders in fact do agree for $n=4$! Of course, this could just be a coincidence, but it is certainly a striking one.
Thoughts?