Consider the following recurrence relation: $$-2a_{n,m} +a_{n-1,m}+a_{n,m-1}=0,$$ where $a_{n,m} \in \mathbb{C}.$I would like a purely combinatorial way to understand the subspace of solutions to this equation which have tempered growth. There is an obvious solution given by setting all $a_{n,m}=C$ for any constant $C,$ and I have reasons (coming from topology) to believe that these are the only solutions with tempered growth.

Now consider the similar but probably much harder recurrence relation: $$-2a_{n,m} +a_{n-1,m}+e^{2 \pi i n \theta}a_{n,m-1}=0$$ where $\theta$ is a fixed irrational. Note that the relation now depends on $n.$ I haven't even been able to come up with a solution to this that has tempered growth. I am hoping that it also has a 1 dimensional (or at least finite dimensional) space of solutions with tempered growth.

Is there a general combinatorial method for attacking either of these recurrence relations? Is there a general way to attack any linear recurrence relation like these?

EDIT: Let me also give my "proof" (I think it is correct) that any solution to the first relation with tempered growth must be constant. Consider the 2 dimensional torus thought of as $S^1\times S^1$, where $S^1$ is the unit circle. Now consider the function $z_1+z_2-2,$ thought of as a (finite) Fourier series. This has only 1 zero, at $(1,1).$

Now consider distributions $D$ on the torus, also thought of as Fourier series $D=\sum_\mathbb{Z^2} a_{n,m}z_1^nz_2^m$ where the $a_{n,m}$ now have only tempered growth. The first recurrence above is exactly the condition that $(z_1,+z_2 -2)D=0$ as a distribution. Since $z_1+z_2-2$ has only a single zero, the only solution with tempered growth should be a multiple of the Dirac distribution which is given by $a_{n,m}=1.$ I want a combinatorial proof or understanding of this phenomenon, since the second relation does not have this kind of topological interpretation. Ideally I would like to prove that the vector space of solutions to the second relation is also dimension 1, or is at least in some way related to the first relation.

2nd EDIT: WillSawin's answer shows that my initial proof is wrong. The space of tempered growth solutions to the first recurrence relation should be spanned (as a vector space) by the delta function and some linear combinations of its partial derivatives. Does the second recurrence have the same property? I.e. is there one "basic solution" $B$ to the second recurrence such that all other solutions can be expressed as linear combinations of the formal derivatives of $B?$