Edit: My original post assumed $p$-arity rather than $p$-regularity. If the $p$ child trees of the root, rather than being (infinite) trees similar to the $p$-regular parent, are of $(p-1)$-arity, then the recursion given needs to be adapted accordingly.
Note however this previous Question and Answer, which appears to give a closed form solution.
The recursion required here is a bit messy but seems to be fairly straightforward.
Let $T_p(m)$ denote the number of rooted (labelled) subtrees of the rooted infinite $p$-arity tree which have $m$ edges and share the same root $v_0$.
Note that the Question asks about an infinite $p$-regular tree, which has arity $p$ for root $v_0$ but all other nodes, having degree $p$, have arity $p-1$. We let $\widetilde{T}_p(m)$ denote this slightly different count and express it in terms of $T_p(m)$.
Essential idea of recursion: Since the root $v_0$ must appear in each subtree, we can choose the number $k$ of the $p$ edges from $v_0$ that will appear in the subtree, and then count possible subtrees extending from those edges.
This gives a recursion on $m$ involving the set $\mathscr{W}(m-k,k)$ of weak compositions of $m-k$ with $k$ summands.
For the basis case, define $T_p(0) = 1$. Then for $m \gt 0$:
$$ T_p(m) = \sum_{k = 1}^{\min(m,p)} \binom{p}{k} \sum_{\vec{w}\in \mathscr{W}(m-k,k)} T_p(w_1)\cdot T_p(w_2) \cdot \ldots \cdot T_p(w_k) $$
Here the inner summation is indexed by weak compositions $\vec{w} = (w_1,w_2,\ldots,w_k)$ of $m-k$ with $k$ summands:
$$ w_1 + w_2 + \ldots + w_k = m-k $$
where the summands are nonnegative integers.
Finally we express the desired $\widetilde{T}_p(m)$ in terms of $T_{p-1}(m)$:
$$ \widetilde{T}_p(m) = \sum_{k = 1}^{\min(m,p)} \binom{p}{k} \sum_{\vec{w}\in \mathscr{W}(m-k,k)} T_{p-1}(w_1)\cdot T_{p-1}(w_2) \cdot \ldots \cdot T_{p-1}(w_k) $$
Added:
Do the cases $m\lt p$ and $m\gt p$ make a difference?
They do in the immediate sense that when $m\gt p$ we are restricted at the root vertex $v_0$ from using up all the edges there (there simply aren't enough to exhaust the $m$ edges of our sought-after subtrees). This shows up in the recursion as the upper limit of the outer summation being given by $\min(m,p)$ rather than depending only on $m$.
