EDIT: I edited this answer rather substantially since I misinterpreted the question a bit at first. Sorry about that.

Formally étale is indeed very weak with respect to what you want:

For example, let me consider the case of a $k$-algebra $A=k[X^{p^{-\infty}}]$ and the map $A \rightarrow k$ given by killing all the variables. This is formally étale but obviously does not have the unique lifting property with respect e.g. to the thickening $A/(x) \rightarrow k,$ which is again given by killing all the variables. A version of this example (with details showing why this is formally étale) is given in this answer.

One candidate for what strongly formally étale might be is what was called *ln-formally étale* last year in M. Morrow's project for the Arizona winter school: Notes with project description, it is project C.1, towards the end of the notes. The "ln" refers to the fact that the lifting property considered there is with respect of killing locally nilpotent ideals. (You can also read there that "ln-formally étale" is an ad hoc term. So does not really answer the question about standard terminology.)

Another candidate is the notion of weakly étale morphism, from B. Bhatt's and P. Scholze's *The pro-étale topology for schemes* (arXiv). A morphism $X \rightarrow Y$ is weakly étale if both itself and its diagonal $\Delta_{X/Y}$ are flat.

It seems that your "strongly formally étale" sits somewhere in between the two (or agrees with one of them):

Clearly strongly formally étale morphism has to be ln-formally étale, since locally nilpotent thickenings are universal homeomorphisms. On the other hand, the argument used in the above linked AWS notes showing "weakly étale $\Rightarrow$ ln-formally étale" should verbatim show that a weakly étale morphism is strongly formally étale: In terms of algebras, if $A \rightarrow B$ is weakly étale, then there is a faithfully flat map $B \rightarrow C$ such that the composite $A \rightarrow C$ is a filtered colimit of étale $A$-algebras (Theorem 1.3 of the Bhatt-Scholze paper linked above). But such algebras have the desired lifting property since étale algebras do, and hence, by fpqc descent of morphisms, so does $A \rightarrow B$.

Unfortunately, it turns out that weakly étale is strictly stronger that ln-formally étale, so there is still some room there. An example demonstrating this is, more or less, the above mentioned example of formally étale algebra, on steroids. A writeup of the example is here.

ADDED: Let me show how to prove, assuming the unique lifting property for weakly étale maps between affine spaces discussed above, that all weakly étale maps have the unique lifting property:

As in the question, assume the commutative square
$$\require{AMScd}(*)\;\;\;\; \begin{CD} Z @>>> Y \\ @VVV \nearrow @VVV\\ W @>>> X \end{CD}$$
where $W \rightarrow Z$ is a universal homeomorphism and $Y \rightarrow X$ is weakly étale. We want to show that the indicated diagonal map, making the whole thing commutative, uniquely exists. (Clearly it uniquely exists as a map of topological spaces since $Z \rightarrow W$ is a homeomorphism - call this map of topological spaces $f$.)

Consider the collection of all commutative squares
$$\require{AMScd}(*')\;\;\;\; \begin{CD} Z' @>>> Y' \\ @VVV @VVV\\ W' @>>> X' \end{CD}$$
such that

1) all the schemes $W'. X', Y', Z'$ are affine open subschemes of $W. X, Y, Z,$ resp., and the maps in the square are obtained by restriction from $(*)$,

2) $Z'$ is the preimage of $W'$, or equivalently, $Z' \rightarrow W'$ is still a universal homeomorphism.

Since the property of $Y \rightarrow X$ being weakly étale is local on the source and the target, it follows that the map $Y' \rightarrow X'$ is weakly étale automatically; thus, assuming (1) and (2), a diagonal map $f': W' \rightarrow Y'$ fitting commutatively into the diagram $(*')$ uniquely exist (and, forgetting the scheme structure, it is clear that this lift as a topological map is obtained by restriction of $f$). So one just needs to glue these maps to a one map $W \rightarrow Z$. To show that this works, one needs the following two claims:

**Claim 1:** Each point $w \in W$ is contained in $W'$ for some square $(*')$ as described above.

**Claim 2:** Given any $w \in W$ and squares $(*'),(*'')$ as above such that $w \in W' \cap W'', f(w) \in Y' \cap Y'',$ there exists another square $(*''')$ as above with $w \in W''' \subseteq W' \cap W'', f(w) \in Y''' \subseteq Y' \cap Y'' $.

Assuming both claims, the lifts $f': W' \rightarrow Y' \hookrightarrow Y$ necessarily glue to the desired map $f: W \rightarrow Y$ (as a map of schemes): Claim 1 makes sure that the map is everywhere defined while Claim 2 implies that the lifts agree on overlaps, because of uniqueness of the lifts (this can always be checked on a basis, in our case basis formed by the affine opens $W'''$ as in the claim). Commutativity of the upper triangle then can be checked again locally, so noting that Claim 1 also implies that the full space $Z$ gets covered (thanks to $Z \rightarrow W$ and all $Z' \rightarrow W'$ being homeomorphisms), the upper of the two triangles commutes, and similarly for the lower triangle. Uniquness follows from uniqueness locally.

So what remains is to check the two claims.
For the first one, consider $w \in W$. Since $Z \rightarrow W$ is a homeomorphism, there is a unique $z \in Z$ mapping to $w$. Denote by $y$ the image of $z$ in $Y$ and by $x$ the image of $w$ in $X$ (so by commutativity of $(*),$ $y \mapsto x$). First choose $x \in X' \subseteq X$ an affine open neighborhood, then $y \in Y' \subseteq Y$ an affine open neighborhood contained in the preimage of $X'$. Next, let $W_1$ denote the preimage of $X'$ in $W$, and let $Z_1$ be the intersection of the preimages of $Y'$ and $W_1$ in $Y$. Since the homeomorphism $Z \rightarrow W$ is open, the image $W_2$ of $Z_1$ in W is an open neighborhood of $w$ contained in $W_1$. Finally, take $x \in W'\subseteq W_2$ an affine open neighborhood, and $Z'$ the preimage of $W'$. Universal homeomorphisms are affine, so $Z'$ is also affine, and hence $W', X', Y', Z'$ assemble to the desired square $(*')$.

The proof of Claim 2 is similar: fixing the same notation for the four points $x, y, z, w$ as above, one sees that $f(w)=y.$ So one can perform the same yoga as for the proof of claim 1, except with the extra condition that $Y'''$ (what was denoted by $Y'$ in the above paragraph) should also be contained in the prescribed intersection $Y' \cap Y''$, and similarly when choosing $W'''$, one has to choose it sufficiently small (inside $W' \cap W''$).

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