Statement. Let $F_X, F_Y$ free groups over $X,Y$ respectively. Suppose there is an isomorphism $\phi: F_X \cong F_Y$. Then $|X|= |Y|$.
My proof. Let $x \in F_X$ be a word of length one, this is, $x = t_1$. Let $\phi(x) = r_1\cdots r_m \in F_Y$. Then $t_1 = \phi^{-1}(r_1)\cdots \phi^{-1}(r_m)$. If $m>1$, there exists $i$ such that $\phi^{-1}(r_i)\phi^{-1}(r_{i+1})=e$, but then $r_ir_{i+1}=e$, in contradiction with $\phi(x)$ being reduced. So $m = 1$. From here is easy to see that $|X \cup X^{-1}| = |Y \cup Y^{-1}|$ and then $|X| = |Y|$.
However, it seems like the proof for the statement is more complicated than this, as in free groups: $F_X\cong F_Y\Rightarrow|X|=|Y|$. So I suppose that my proof is wrong, but I don't know where.
Consider the automorphism of the free group on two elements, $x$ and $y$, given by conjugation by $y$. So $\phi\colon F_2\to F_2$ is an isomorphism, and $\phi(x)=yxy^{-1}$, $\phi(y)=y$. We can see that your conclusion that $\phi(x)$ must have "length $1$" in the basis of the image is false, so running this example through your proof will identify the first flaw in your argument.
Here, $\phi^{-1}(y) = y$, and $\phi^{-1}(x) = y^{-1}xy$ (since the inverse of conjugation by $y$ is conjugation by $y^{-1}$).
So your argument goes:
(because you say that one of the two consecutive terms must cancel)
But that does not hold: $\phi^{-1}(y) = y$, $\phi^{-1}(x)=y^{-1}xy$, so $\phi^{-1}(y)\phi^{-1}(x) = xy$, and $\phi^{-1}(x)\phi^{-1}(y^{-1}) = y^{-1}x$. In either case, they do not fully cancel as you claimed they "should".
So the first error occurs in your claim that if $m\gt 1$ there must exist $i$ such that $\phi(r_i)\phi(r_{i+1})$ cancel out completely.
Note that , as I mentioned at the end in my answer to the post you referenced, you can have an automorphism of $F_2$ that sends elements of $\{x,y\}$ to elements of pretty much any length, by composing the map $x\mapsto xy$ and $y\mapsto y$; $x\mapsto x$ and $y\mapsto xy$; and conjugations. So this kind of argument is doomed.