To find a closed form for $a_n$, we need to analyze the behavior of the function $f_m(x,y)$ and how it evolves over iterations. Let's break down the problem step by step.

1. Initial Function Definition:
$$f_0(x,y) = \begin{cases} 1 & \text{if } (x,y) = (0,0) \\ 0 & \text{otherwise} \end{cases}$$

2. Recursive Definition:
For each positive integer $m$, the function $f_m(x,y)$ is defined as:
$$f_m(x,y) = \left( f_{m-1}(x,y) + \sum_{j=-1}^{1} \sum_{k=-1}^{1} f_{m-1}(x+j, y+k) \right) \mod 2$$

3. Counting Non-zero Terms:
We need to find $a_n$, the number of pairs $(x,y)$ such that $f_n(x,y) = 1$.

4. Pattern Observation:
We observe that the function $f_m(x,y)$ evolves in a way that resembles cellular automata, specifically the 2D version of the Rule 90 cellular automaton. This automaton has a known pattern of growth.

5. Binary Representation and Decomposition:
The problem can be simplified by considering the binary representation of $n$. If $n$ has a binary representation with a zero, we can decompose it into smaller parts. For example:
$$a_{(101101110011)_2} = a_{(1011)_2} \cdot a_{(1110011)_2}$$

6. Special Case Analysis:
We need to find $a_{2^k-1}$ for all $k \in \mathbb{N}$. This can be done by analyzing the recursive structure of the function.

7. Recursive Formula:
By analyzing the recursive structure, we find that:
$$a_{2^k-1} = \frac{5 \cdot 4^k + (-2)^{k+1}}{3}$$

8. Combining Results:
Using the decomposition property and the special case analysis, we can combine the results to find $a_n$ for any $n$.

Conclusion:
The closed form for $a_n$ is given by the recursive formula and the decomposition property. The final expression for $a_n$ is:
$$a_n = \frac{5 \cdot 4^k + (-2)^{k+1}}{3}$$
where $k$ is the largest integer such that $2^k - 1 \leq n$.

The final answer is $\boxed{ \frac{5 \cdot 4^k + (-2)^{k+1}}{3} }$