It is a very nice question!
Theorem. $\Delta(\omega_2,\omega_1)$ is equivalent to CH.
Proof: You've pointed out that CH implies the principle, since the hypothesis you mention for this case amounts to $\omega_1^{\lt\omega_1}<\omega_2$, which amounts to CH. So let us consider what happens when CH fails. Let $T=2^{\lt\omega}$ be the tree of all finite binary sequences, and label the nodes of $T$ with distinct natural numbers. Let $F$ be the subsets of $\omega$ arising as the sets of labels occuring on any of $\omega_2$ many branches through $T$. Thus, $F$ has size $\omega_2$, and any two elements of $F$ have finite intersection. I claim that this family of sets can have no $\Delta$-system of size $\omega_2$, and indeed, it can have no $\Delta$-system even with three elements. If $r$ is the root of $a$, $b$ and $c$ in $F$, then $r=a\cap b=a\cap c$, and so $a$ and $b$ branch out at the same node that $a$ and $c$ do, in which case $b$ and $c$ must agree one step longer, so $b\cap c\neq r$. QED
The same idea works for higher cardinals as follows:
Theorem. For any infinite cardinal $\delta$, we have $\Delta(\delta^{++},\delta^+)$ is equivalent to $2^\delta=\delta^+$.
Proof. If $2^\delta=\delta^+$, then your criterion of $(\delta^+)^{\lt\delta^+}<\delta^{++}$ is fulfilled, and so the $\Delta$ property holds. Conversely, consider the tree $T=2^{\lt\delta}$. Label the nodes of this tree with $2^{\lt\delta}$ many distinct labels (perhaps this is more than $\delta$, but it is no matter). Let $F$ be a family of $\delta^{++}$ many sets arising as the label sets of this many branches through $T$. Each of these sets has size $\delta$, since a branch has $\delta$ many labels. But again, for the same reason as above, there can be no $\Delta$ system with even three elements, since the tree is only binary branching, and this contradicts $\Delta(\delta^{++},\delta^+)$. QED