Let $\mathcal L_Q$ be the logical system that includes first order logic together with the quantifier $Q$ which is defined as follows:
For an interpretation $\mathfrak I=(\mathfrak A, \beta)=((A,\mathfrak a), \beta)$,
$\mathfrak I\models Qx\varphi$ iff there exist uncountably many $a\in A$ so that $\mathfrak I \frac a x\models \varphi$.
My question is the following:
Let $\Phi$ be a set of formulas satisfied by a model of cardinality $\geq \aleph_1$, is $\Phi$ satisfiable by a model of cardinality $>\kappa$ for any cardinality $\kappa$?
If the answer to the above is negative, does the weaker version where $\kappa=\mathfrak c$ have a positive answer?
No, it doesn't.
Consider $\omega_1$, thought of as a linear order. This satisfies - in addition to the usual axioms of linear order - the axioms "The universe is uncountable" and "For each $x$, the set $\{y: y<x\}$ is countable." But every uncountable linear order with the countable predecessor property has cardinality $\aleph_1$.
What about $\aleph_2$?
Well, we can use a variant of the same trick! Consider the structure defined as follows:
We start with $\omega_2$ as a linear order.
Now we attach to each $x\in\omega_2$ a copy of $\omega_1$ as a linear order, together with a surjection from that copy to the initial segment $\{y: y<x\}$ of $\omega_2$. We now have a two-sorted structure, one of the sorts being the $\omega_2$-part and the other sort being the disjoint union of a bunch of $\omega_1$s, and some stuff connecting them appropriately.
Any $\mathcal{L}_Q$-elementarily equivalent structure must have, in place of the $\omega_2$-part, a linear order with the $\le\aleph_1$-predecessor property by the argument above for $\omega_1$. But this rules out models of cardinality $\aleph_3$ or above.
Similarly we can whip up a theory with models of size $\aleph_4$ but no models of size $\ge\aleph_5$, and so on. And in fact we don't need theories: all of this can be accomplished by individual $\mathcal{L}_Q$-sentences.
The first point where things become a bit slippery is $\aleph_\omega$; if I recall correctly we can continue well past this, but I don't immediately see the details.