Let $A$ be a set and $n,m\in\mathbb{N}$. Suppose we have an $n$-tuple $(a_1,\ldots,a_n)$ and an $m$-tuple $(b_1,\ldots,b_m)$ of elements of $A$. Without assuming anything about the relation between $n$ and $m$, does the equality $$(a_1,\ldots,a_n)=(b_1,\ldots,b_m)$$ imply $n=m$? What happens if $n\ne m$? Surely that would not give a contradiction since $(a_1,\ldots,a_n)\ne(b_1,\ldots,b_m)$ would be just as ridiculous for $n\ne m$.
This problem doesn't occur with finite sequences: i.e. with members of $\bigcup_{n\in\mathbb{N}} A^{[1,n]}$. If $x,y\in\bigcup_{n\in\mathbb{N}} A^{[1,n]}$ and $x=y$, then domain of $x$ must be equal to the domain of $y$, since we are dealing with an equality of functions. For example, if $x:[1,n]\rightarrow A$ and $y:[1,m]\rightarrow A$ and $x=y$, then by definition $n=m$. Is it the same case for tuples?
You say "this problem doesn't occur with finite sequences", but $n$-tuples ARE finite sequences. Very commonly, they are defined as such. Even if they are given some different set-theoretic construction, the $n$-tuple with entries $a_i$ for $i \in \{1,\dotsc,n\}$ can be identified with the function $i \mapsto a_i$.
(As far as I've been taught, every indexed set is a function; $n_i$ is just an alternative notation for $n(i)$.)
Of course, if $n \neq m$, then any $n$-tuple $(a_1, \dotsc, a_n)$ is not equal to any $m$-tuple $(b_1, \dotsc, b_m)$. This should be included in any definition of equality of tuples.
If $n = m$, then the tuples are equal if $a_i = b_i$ for each $i \in \{1, \dotsc, n\}$.