I think it is true. Moreover I think you can replace $2^\omega$ with any T1 topological space, and it's true for $n$-bushy trees with $4^\omega$ replaced by $(3n-2)^\omega$.
For a $2$-bushy tree, let's call the node $\sigma$ the start of the tree. (I'd call it the root, but you might want all trees to be rooted at the empty sequence.)
Suppose there are no $2$-bushy trees $T$ starting at the root of $4^{\leq \omega}$ such that $f$ restricted to $[T]$ is one-one. Call a tree $T$ $k$-injective if any two branches through $T$ that get mapped by $f$ to the same point agree for the first $k$ nodes. Then a limit of a sequence $T_k$ of $k$-injective $2$-bushy trees starting at the root is a $2$-bushy tree on which $f$ is injective.
Hence for some $k$, there are no $k$-injective $2$-bushy trees starting at the root. Make a $2$-bushy tree starting at the root that is arbitrary in the first $k$ levels. There must be no way of continuing this to a $k$-injective tree. Hence we have $2^k$ nodes such that, for every $2^k$-tuple of $2$-bushy trees starting at those nodes, a branch from two of them must map to the same point.
Lemma: Let $T$ be a $a$-bushy tree starting at a node $v$. Let $Z$ be a closed set of $2^{ <\omega}$. Suppose that, for every $b$-bushy subtree $T'$ of $T$ starting at $v$, $f([T']) \cap Z \neq \varnothing$. Then there is an $a-b+a$-bushy subtree $T^*$ of $T$ such that $f(T^*) \subseteq Z$.
Proof: Let $T^*$ be the subtree of $T$ consisting of all vertices that are not the start of an $b$-bushy subtree whose branches all lie in $2^{ \omega}-Z$. Then every node of $T^*$ has at least $a+b-1$ successors in $T^*$ - if it had any fewer, it would have $b$ sucessors, each the start of a $b$-bushy subtree whose branches lie in $2^{\omega}-Z$, hence it would be the start of a $b$-bushy subtree whose branches lie in $2^{\omega}-Z$. $T^*$ contains $v$. Finally $[T^*] \subseteq Z$ because any branch not in $Z$ would have an open neighborhood of branches not in $Z$, in which it would be easy to form a $b$-bushy tree starting at one of the vertices of the branch whose branches do not intersect $Z$.
Now pick a node and let $Z_1$ and $Z_2$ be two closed sets such that, for every $2$-bushy tree $T$ starting at that node, $f([T])$ intersects $Z_1$ and $Z_2$.