This is a partial answer to this question:
I was always under the impression that a type level pair must depend on having some kind of well ordered infinite class of objects, well at least this was the experience I found with the Quine-Rosser pair, and in addition it should presuppose Extensionality (like Quine-Rosser pairs, or even Holmes's $1$-high type pairs) However. It's only today when I came to realize that this is not necessarily correct.
I think I can cook up a pair that fulfills the aforementioned criteria.
Lets work in $\sf ZFA \neg C$.
Now, let $F_1$ be a total injective function that sends sets to nonempty sets of nonempty sets not having $1$ among their elements. That is: $$ F_1(x)=y \to \exists z\, (z \in y) \land \forall z (z \in y \to \exists u \, (u \in z) \land 1 \not \in z)$$ Similarity let $F_2$ have the same above properties but with respect to $2$. So: $$ F_2(x)=y \to \exists z\, (z \in y) \land \forall z (z \in y \to \exists u \, (u \in z) \land 2 \not \in z)$$
Now, we define an "inserter" function $I_i$ that inserts $i$ to all elements of a set, that is:
$$I_i(x)=\{y \cup \{i\} \mid y \in x \}$$
Now we define the following pair:
$$(a,b) = I_1 (F_1(a)) \cup I_2 (F_2(b))$$
We can easily retrieve the $i^{th}$ projection of $(a,b)$: We take the set of all elements of $(a,b)$ that have $i$ among their elements, apply the de-inserter function $I_i^{-1}$ on it, then apply $F_i^{-1}$ and we get the $i^{th}$ projection of $(a,b)$.
We can extend that to any $\lambda$-tuple: $$(x_1,x_2,...)^\lambda = \underset {\alpha < \lambda} \bigcup I_\alpha(F_\alpha(x_\alpha))$$
Of course, to answer the above question, we can simply take the tuple to be: $$\{\{(x_1,x_2,...)^\lambda\}\}$$ The main drawback is that this definition is not that general, for instance it doesn't work in the usual known models of $\sf NFU$, where we have strictly more empty objects than nonempty. In such models I only know of pairs with height of at least $2$ that can do the job.