You can use generating functions, as we did in the single variable case.
Let $G(x,y)=\sum_{m,n\ge 0}F(n,m) x^n y^m$. We'll express $G$ in a nice form from which one can recover $F(n,m)$.
As you didn't specify initial conditions, let
$$H_1(x)=\sum_{n\ge0} F(n,0)x^n, H_2(y)=\sum_{m\ge0} F(0,m)y^m, c=F(0,0)$$
By the recurrence of $G$, if we multiply it by $1-x-y$, most of the terms will cancel. I'll elaborate on that.
I choose $1-x-y$ in a similar manner to that of constructing the characteristic polynomial in one variable: $1$ corresponds to $F(n,m)$, $x$ to $F(n-1,m)$ and $y$ to $F(n,m-1)$, i.e. $F(n-a,m-b)$ is replaced by $x^ay^b$.
$$G(x,y)(1-x-y)=\sum_{m,n\ge 0}F(n,m) (x^n y^m-x^{n+1}y^m-x^{n}y^{m+1})=$$
We'll group coefficients of the same monomial:
$$\sum_{m,n \ge 1} (F(n,m)-F(n-1,m)-F(n,m-1)) x^{n}y^{m}+$$
$$\sum_{n \ge 1} (F(n,0)-F(n-1,0)) x^{n}+\sum_{m \ge 1} (F(0,m)-F(0,m-1)) y^{m}+F(0,0)=$$
$$H_1(x)(1-x) + H_2(y)(1-y)-c$$
So, finally,
$$G(x,y) = \frac{H_1(x)(1-x) + H_2(y)(1-y)-c}{1-x-y}$$
(Compare this to the relation $Fib(x)=\frac{x}{1-x-x^2}$ where $Fib$ is the generating function of the Fibonacci sequence.)
How do we recover $F$? We use the formal identity $\frac{1}{1-x-y}=\sum_{i\ge 0}(x+y)^i$. Let $S(x,y)=H_1(x)(1-x) + H_2(y)(1-y)-c=\sum_{n,m} s_{n,m} x^ny^m$.
It gives us:
$$G(x,y) = \sum_{i \ge 0}S(x,y)(x+y)^i = \sum_{n,m \ge 0} (\sum_{a,b \ge 0}s_{a,b} \binom{n+m-a-b}{n-a})x^ny^m$$
So $F(n,m) = \sum_{a,b \ge 0}s_{a,b} \binom{n+m-a-b}{n-a}$. I have an hidden assumption - that $S$ is a polynomial! Otherwise convergence becomes an issue.
I guess that your initial conditions are $F(n,0)=1, F(0,m) = \delta_{m,0}$, which give $S(x,y)=1$, so $F(n,m)=\binom{n+m}{n}$.
EDIT: In the general case, where $F(n,m)=\sum_{a,b} c_{a,b}F(n-a,m-b)$ where the sum is over finitely many tuples in $\mathbb{N}^{2} -\setminus \{ (0,0) \}$, the generating function will be of the form $\frac{H(x,y)}{1-\sum_{a,b} c_{a,b}x^a y^b}$ where $H$ depends on the initial conditions.
When we had one variable, we wrote $\frac{q(x)}{1-\sum a_i x^i} =\sum \frac{q_i(x)}{1-r_i x}$ where $r_i^{-1}$ is a root of $1-\sum a_i x^i$ and used $\frac{1}{1-cx} = \sum c^ix^i$.
With 2 variables, this is not always possible, but we can write $\frac{1}{1-\sum_{a,b} c_{a,b}x^a y^b}=\sum_{i \ge 0} (\sum_{a,b} c_{a,b}x^a y^b)^{i}$ and use the binomial theorem to expand. We can also use complex analysis methods to derive asymptotics of $F(n,m)$ from the generating functions.
Best Answer
This looks right - it would help clarity if you gave the actual recurrence you are working with - I assume it is $a_{n+4}=10a_{n+3}-37a_{n+2}+60a_{n+1}-36a_n$
Have you tried checking that your proposed solution works? Since four items determine the whole thing, you need the four coefficients, but since this is linear you could check $2^n, n2^n, 3^n$ and $n3^n$ separately.