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