A clutter is a pair $C=(V,E)$ where $V\neq\emptyset$ is a set, and $E\subseteq {\cal P}(V)$ such that no member of $E$ is included in another member of $E$. A matching in $C$ is a collection of pairwise disjoint members of $E$. Zorn's Lemma implies that every matching is contained in a maximal matching (with respect to $\subseteq$).
Is it possible to find a clutter $C=(V,E)$ with $E\neq\emptyset$ and for every matching $M\subseteq E$ there is a matching $M'\subseteq E$ with $|M|<|M'|$?
Yes, this is possible. For each prime $p$ and $c \in \{0,1, \dots, p-1\}$ let $A_{c,p}=\{c+kp \mid k \in \mathbb{Z}\}$. Clearly, the set of all $A_{c,p}$ is a clutter $\mathcal C$ with ground set $\mathbb{Z}$. If we fix $p$, then the set of $A_{c,p}$ is a matching of size $p$. Since there are infinitely many primes, $\mathcal C$ has arbitrarily large finite matchings. On the other hand, no matching of $\mathcal C$ is infinite. To see this, it suffices to prove that if $A_{c_1,p_1}$ and $A_{c_2,p_2}$ are disjoint, then $p_1=p_2$. Suppose $p_1 \neq p_2$. By shifting both sequences by $c_1$, we may assume that $c_1=0$. Choosing $k \equiv -c_2 p_2^{-1} \pmod{p_1}$, we have $c_2+kp_2 \equiv 0 \pmod{p_1}$, and so $A_{c_1,p_1} \cap A_{c_2,p_2}$ is non-empty.
In the same spirit of Tony Huynh's answer:
Take $V = \prod_{n < \omega} n + 1 = \{f \colon \mathbb{N} \to \mathbb{N} \mid \forall n,\, f(n) \leq n\}$. Take $A_{n, i} = \{f \in V \mid f(n) = i\}$. Take $E = \{A_{n,i} \mid n < \omega, i \leq n\}$. Clearly, $A_{n,i} \cap A_{m,j} = \emptyset$ iff $n = m, i \neq j$. Thus, the sets of pairwise disjoint elements from $E$ are of any finite size, while $E$ is infinite. By a slight modification of the same idea we get:
Claim: Let $\mathcal{K}$ be an infinite collection of non-zero cardinals. There is a set $X$ of size $\sup \mathcal{K}$, a collection $E \subseteq P(X)$ of size $\prod \mathcal{K}$ such that for every $a, b \in E$, $a \not\subseteq b$ and the collection of cardinalities of maximal pairwise disjoint subsets of $E$ is exactly $\mathcal{K}$.
Proof: Take $X$ to be the finite support product of $\mathcal{K}$ (namely, choice functions for $\mathcal{K}$ which are non-zero only for finitely many coordinates). For every $x \in \prod \mathcal{K}$ (the full support product) which is identically zero at most once, and for every $\kappa \in \mathcal{K}$ and $\alpha \in \kappa$, take
$$A_{\kappa, \alpha, x} = \{f \in X \mid f(\kappa) = \alpha,\, \exists \mu \in \mathcal{K}, f(\mu) \neq 0, x(\mu)\},$$ and $E$ the collection of all $A_{\kappa,\alpha, x}$ such that $x(\kappa) = \alpha$. For $\kappa \neq \kappa'$ in $\mathcal{K}$, $\alpha \in \kappa, \alpha' \in \kappa'$ and $x, x'$ take $f$ to be the function such that $f(\kappa) = \alpha,\, f(\kappa') = \alpha'$ and pick one additional elements in $\mathcal{K}$ of size $>2$ to conflict with both $x$ and $x'$. Then $f \in A_{\kappa, \alpha, x} \cap A_{\kappa', \alpha', x'}$.
This argument shows that if two sets in $E$ are disjoint then $\kappa = \kappa', \alpha \neq \alpha'$, giving an explicit description of all maximal pairwise disjoint subsets of $E$.
We need to show that for all $a, b \in E$, $a \not\subseteq b$. Take $a = A_{\kappa,\alpha, x}, b = A_{\kappa', \alpha', x'}$. If $\kappa \neq \kappa'$ or $\alpha \neq \alpha'$, take $f(\kappa) = \alpha$, $f(\kappa') \neq \alpha'$. Otherwise, take $\mu \in \mathcal{K}$ such that $x(\mu) \neq x'(\mu)$ and set $f(\mu) = x'(\mu)$. Set all other values of $f$ to be zero, then $f \in A_{\kappa,\alpha, x}$ but $f \notin A_{\kappa',\alpha', x'}$.
