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 $f \notin A_{\kappa',\alpha', x'}$.