My goal is to obtain the Big-Oh bound of the following recursive function with two variables:

$$T(n,m) = T(n, m-1) + T(n-1,m)+1$$

As initial conditions, $T(0,m)=1$ and $T(n, 0)=1$ for $m \geq 0$ and $n \geq 0$, respectively. Then, I think $T(n,m) \in O(2^{n+m})$ which can be proved as follows:

Proof:
* Let's prove $T(n, m) \leq 2 \times 2^{n+m} - 1$ by induction.
   - Base cases: $T(0, m) = 1 \leq 2 \times 2^{m} -1$ and $T(n, 0) = 1 \leq 2 \times 2^{n} -1$ for $n,m \geq 0$
   - Inductive step
      * Assume $T(k, p) \leq 2 \times 2^{k+p} - 1$ for arbitrary $k$ and $p$. Then, there two next steps: $T(k+1, p)$ and $T(k, p+1)$.
      * Case 1) $T(k+1, p) = T(k+1, p-1) + T(k, p) + 1$ by the recursion. Using the above assumption, $T(k+1, p) \leq 2\times 2^{k+p+1} - 1$; thus, $T(k+1, p)$ holds true.
      * Case 2) $T(k, p+1)$ also holds similarly to Case 1.
   - Hence, $T(n, m) \leq 2 \times 2^{n+m} - 1 \in O(2^{n+m})$ . 

My questions are
* Are there errors in the above proof?
* Are there tighter bounds than $O(2^{n+m})$?