Since all vertices are reachable from $1$, we can select $n-1$ edges forming a directed tree rooted at $1$. It has exactly one path to each vertex.

Therefore, if $m \le n$, deleting at most one edge leaves such a tree, and the answer is $n$.

Otherwise, compute a topological order. Let $z$ be the first vertex in this order with at least two incoming edges. Every earlier vertex has a unique path from $1$: except for $1$, it has exactly one predecessor, which is also earlier.