Can we have an infinite sequence of decreasing cardinality all terms of which have equal sized power sets? Is the following consistent with $\text{ZF}$? 
There exists a set $S=\{x_1,x_2,x_3,...\}$ such that:


*

*$|x_{i+1}|  < |x_i|$ 

*$\forall m,n \in S (|P(m)|=|P(n)|)$
Where cardinality $``||"$ is defined after Scott's.
 A: This is consistent, at least under a rather tame large cardinal assumption. (One can also produce examples by manipulating Dedekind finite sets, but Asaf's answer addresses this. The answer here works even in the context of $\mathsf{DC}$.)
For instance, see

MR3612001. Conley, Clinton T.; Miller, Benjamin D. Measure reducibility of countable Borel equivalence relations. Ann. of Math. (2) 185 (2017), no. 2, 347–402. 

There, Clinton and Ben show that every basis for the nonmeasure-hyperfinite countable Borel equivalence relations under measure reducibility is uncountable. A basis here is a set $B$ such that given any such equivalence relation, there is one below it (in the ordering of measure reducibility) and in $B$. They explain how their arguments give stronger results, for instance, continuum-many pairwise incomparable such relations, or infinite strictly decreasing sequences. They also explain how in $L(\mathbb R)$, under the assumption of determinacy, their results actually give that the corresponding quotients $|\mathbb R/E|$ are decreasing in cardinality.
I'm fairly certain that this result (the existence of such decreasing sequence of equivalence relations or, under determinacy, of such a decreasing sequence of cardinals) predates the Conley-Miller paper by more than 10 years, but had a bit of trouble tracking a specific reference. The point is that incomparability results in the theory of countable Borel equivalence relations are typically established via Baire category arguments, so they hold for true cardinality in, say, $L(\mathbb R)$ if determinacy holds, since then all sets of reals have the Baire property. 
Anyway, the other point is that all the relations under consideration here are such that $|\mathbb R|<|\mathbb R/E|$ (in fact, $|\mathbb R/E_0|<|\mathbb R/E|$, where $E_0$ is the Vitali equivalence relation), so we also have that the power sets of all these cardinals have the same size $2^{\mathfrak c}$, by the argument indicated in the answers to this previous question.
A: Yes.
For silly reasons.
Suppose that $X$ is a Dedekind-finite set, then $S(X)$, the set of all injective finite sequences from $X$ is also Dedekind-finite. Let $S_n(X)$ denote the subset of $S(X)$ of sequences whose domain is at least $n$. It is easy to see why $S_n(X)$ surjects onto $S(X)$. Simply erase the first $n$ coordinates.
This means, by arguments from my answer to your last question, that $S(X)\leq^* S_n(X)$ for all $n$. Since those are Dedekind-finite sets, and the inclusion is strict, they form the wanted sequence.
You can even be more clever than this, and for some chain in $\mathcal P(\omega)$ which has order type $\Bbb R$, define a sequence of order type $\Bbb R$ of Dedekind-finite sets, all of which have equipotent power sets.
