Let $\mathcal{T}$ be a first-order theory with arbitrarily large finite models. It is true that $\mathcal{T}$ has models of every infinite cardinality? If the signature is countable, the answer is yes by the compactness theorem and Löwenheim–Skolem theorem. But if the signature is uncountable, it's unclear to me what happens for infinite cardinalities smaller than the size of the signature.
Logic – Upward Löwenheim–Skolem Theorem for Uncountable Signatures
logicmodel-theory
Best Answer
Complementing Greg Nisbet's answer, here's a proof that $\beth_1$ is the best one can do (this is a quick modification of the usual example of a countably infinite structure with no countable proper elementary extension.):
Let $A_n$ be the linear order with $n$ elements. Let $B_n$ be the binary tree of height $n$, viewed as a structure in the language consisting of a constant $o$ naming the root, two successor relations $S_0$ and $S_1$ (left and right successor, respectively), and the "is a prefix of" relation. For each $r\in 2^\mathbb{N}$, let $U_r$ be a unary relation symbol. Finally, let $E$ be a binary relation symbol and consider the following two-sorted theory $T$:
The common theory of the $A_n$s holds on the first sort and the common theory of the $B_n$s holds on the second sort.
Via $E$, the first sort "counts" the levels of the second sort. E.g. $o$ is the unique thing which $E$ pairs with the smallest element of the first sort, and the children of $\sigma$ get $E$-paired with the successor of the thing $\sigma$ gets $E$-paired with.
Each $U_r$ names a maximal chain in the tree: if $U_r(\sigma)$ and $U_r(\tau)$ then $\sigma\preccurlyeq\tau$ or $\tau\preccurlyeq\sigma$, and each point on the first sort gets $E$-paired with some $\sigma$ with $U_r(\sigma)$.
For each $r\in 2^{\mathbb{N}}$ and each finite binary string $\sigma\prec r$, the axiom "If $\sigma$ exists, then $U_r(\sigma)$." Here "$\sigma$ exists" is shorthand, expressed in terms of the successor relations; e.g. "$0110$ exists" is really $$\exists x_1,x_2,x_3,x_4(oS_0x_1\wedge x_1S_1x_2\wedge x_2S_1x_3\wedge x_3S_0x_4),$$ and $x_4$ is the node corresponding to $0110$.
Now any infinite model of the above theory must have continuum-many elements. One key point is that since all the $A_n$s have a greatest element, the full binary tree of height $\omega$ is not (the tree part of) a model of this theory. Once we go infinite at all, we have to go all the way to $\beth_1$.