What is the solution of $T(n, m) = T(n, m-1) + T(n-1, m) + c$?
The function $f(n,m) = \binom{n+m}{n}$ satisfies the recurrence
$$
f(n,m) = f(n-1,m) + f(n,m-1),
$$
with base cases $f(n,0) = f(0,m) = 1$. This already gives you the solution when $c = 0$, assuming that the base cases are $f(n,0) = f(0,m) = 1$.

One function which satisfies the recurrence, though (in general) not the base case, is $g(n,m) = -c$:
$$ g(n,m) = g(n,m-1) + g(n-1,m) + c. $$
This gives the solution when $c = -1$.

Finally, note that any function of the form $\alpha f(n,m) + g(n,m)$ is also a solution of the recurrence. Since
$$
\alpha f(n,0) + g(n,0) = \alpha f(0,m) + g(0,m) = \alpha - c,
$$
we see that
$$
T(n,m) = (c + 1) f(n,m) + g(n,m) = (c+1) \binom{n+m}{n} - c.
$$