Conjecture that Wilf equivalence implies enumerating equivalence for binary-tree patterns

Let $s$ and $t$ be patterns on binary trees. Two patterns are Wilf equivalent when their avoidance functions are equal, and they are enumerating equivalent when their enumerating generating functions are equal. The avoidance function of a pattern $t$ is $Av_t(x)=\sum_{n\geq 1}a_nx^n$, while its enumerating generating function is $En_t(x,y)=\sum_{n\geq 1,\,k\geq 0}a_{n,k}x^ny^k$, where $a_n$ counts binary trees avoiding $t$ and $a_{n,k}$ counts those containing exactly $k$ copies of $t$. Wilf-to-enumerating equivalence conjecture. If two patterns $s$ and $t$ are Wilf equivalent, then they are enumerating equivalent. Since $Av_t(x)=En_t(x,0)$, enumerating equivalence is a stronger condition than Wilf equivalence; the conjecture asserts that these two notions nevertheless coincide for such patterns.