Do all single variable identities over exponentiation have the same length? Consider exponentiation on the natural numbers. Suppose we have an equation involving only the variable $x$, where there are $m$ $x$'s on one side and $n$ $x$'s on another side where $m$ and $n$ are different, like $(x^x)^{x^x}=(x^x)^{(x^x)^x}$, where the brackets can go anywhere. Is it always the case that there is a natural number such that the equation does not hold? In other words, do all single variable identities over exponentiation have the same length?

Yes your claim is true, and actually we can do much better. If $f(x)$ and $g(x)$ are two exponential functions of the form you ask about, and they aren't the same, then one of them asymptotically dominates the other.
My proof is similar to Matija's but a bit shorter, since we don't need to invoke the whole structure of binary expression trees for this problem; thus we can define a simpler tree-to-exponential expression correspondence.
Write $f\gg g$ if $\lim\limits_{x\rightarrow\infty} \frac{f(x)}{g(x)} = \infty$. We make use of two simple facts:
Fact 1: $f\gg g$, and $\liminf\limits_{x\rightarrow\infty} g(x)>0$, then $x^{f(x)} \gg x^{g(x)}$.
Fact 2: There is a one-to-one correspondence between functions formed by exponentiating $x$'s and unordered rooted trees.
The correspondence goes like this: Let $f_T(x)$ represent the function corresponding to tree $T$. If $T$ is a single node, then $f_T(x) = x$. If $T$ has subtrees $T_1,T_2,\dots,T_k$ attached to its root, then we have $$
f_T(x) = ((x^{f_{T_1}(x)})^{f_{T_2}(x)})\cdots)^{f_{T_k}(x)}
$$
The fact that ordering on the nodes is irrelevant corresponds to the algebraic fact that $(a^b)^c =(a^c)^b$. It will also be convenient to write the empty tree as $0$, with $f_0(x) = 1$.
For example, look at these trees:

We have $$
F_A(x)=x^{((x^x)^x)^x}\hspace{8pt}F_B(x) = (x^x)^x\hspace{8pt}F_C(x)=x^{x^{x^x}}\hspace{8pt}F_D(x) =x^{\left(\left((x^x)^{x^x}\right)^{(((x^x)^x)^x)^x}\right)}
$$
We define a total order on tree inductively as follows: $T>S$ if $S=0$ and $T\ne 0$. If $T$ and $S$ are nonempty, and have subtrees attached to their root $T_1,T_2,\dots$ and $S_1,S_2,\dots$ respectively (all but finitely many of $T_i$ and $S_i$ are $0$) and these are ordered in decreasing order (i.e. $T_1\ge T_2\ge \cdots$ and the same for $S$), then we say $T>S$ if the tuple $(T_1,T_2,\dots)>(S_1,S_2,\dots)$ in lexicographic order.
Theorem: If $T>S$, then for all $p\in (0,\infty)$ $$
f_T(x) \gg f_S(x)^p
$$
In particular $f_T(x) = f_S(x)$ has finitely many solutions in $\mathbb N$ if $T\ne S$. The number of nodes of a tree corresponds to the number of $x$'s in the expression of $f_T(x)$, hence your conjecture is true.
Proof: We induct on trees. If $S=0$, this is trivial. Otherwise, suppose $T>S$ have $(T_1,T_2,\dots)$ and $(S_1,S_2,\dots)$ are the subtrees meeting at the root of $T$ and $S$, indexed in decreasing order; we show that if the claim holds for all $T_i$ and $S_i$ then it holds for $T$ and $S$. By $T>S$, there exists $k$ such that $T_k>S_k$ and $T_i=S_i$ for $i<k$. Observe, for any $p\in(0,\infty)$
$$
\lim\limits_{x\rightarrow\infty}\frac{\prod_{i\in\mathbb N} f_{T_i}(x)}{p\prod_{i\in\mathbb N} f_{S_i}(x)} = \frac1p\lim\limits_{x\rightarrow\infty}\frac{f_{T_k}(x)}{f_{S_k}(x)}\frac{\prod_{i>k} f_{T_i}(x)}{\prod_{i>k} f_{S_i}(x)} \ge \frac1p\lim\limits_{x\rightarrow\infty}\frac{f_{T_k}(x)}{f_{S_k}(x)^N} = \infty
$$
where $N$ is the number of nonzero trees in $\{S_i\}$. Noting that the numerator and denominator of the fraction on the RHS are $\log_x(f_T(x))$ and $\log_x(f_S(x)^p)$ respectively, fact 1 implies the result.
