I claim that the only possibility is $m=2^\ell-1$, $n=1$. First note that there are $2^\ell-1$ subsets of $\{1,2,\cdots,\ell\}$, and we generate $mn$ sets of the form $ A_{i}\Delta B_{j} $. Thus $mn=2^\ell-1$. To see that $m=2^\ell-1$, $n=1$ is possible, let $B_1=\oslash$ (the empty set), and let $A_1, A_2, \ldots, A_{2^\ell-1}$ be any permutation of the nonempty subsets of $ \{1,2,\cdots,\ell\} $. Then $A_i \Delta B_1=A_i$, so indeed we generate all the nonempty subsets required. Assume that $n \geq 2$. In the next 3 paragraphs we shall show that we can assume that the empty set is not one of $A_1, A_2, \ldots, A_m, B_1, B_2, \ldots, B_n$. We shall define an operation on our sets $A_1, A_2, \ldots, A_m, B_1, B_2, \ldots, B_n$. Let $O(i)$ be the operation that for each set $X$ mentioned in the last sentence, if $i \in X$ then remove it, and if $i \notin X$ insert $i$ into the set $X$. Note that our new set is equal to $X \Delta {i}$. Considering $A_{i}\Delta B_{j} $, we easily see that doing $O(i)$ to all the sets does not change this. Thus we may perform as many operations of the form $O(i)$ as we like to our sets while still keeping the property that their symmetric differences generate all nonempty subsets of $\{1,2,\ldots, \ell\}$. I now claim there is a set $S\subset \{1,2,\ldots, \ell\}$ with $S$ being distinct from all $A_1, A_2, \ldots, A_m, B_1, B_2, \ldots, B_n$. This amounts to showing that $m+n<2^\ell-1$. Well, $m+n=n+\dfrac{2^\ell-1}{n}$. Taking the derivative with respect to $n$ and seeing that since $m \geq n$, then $\sqrt{2^\ell-1} \geq n$, so our function is decreasing in $n$. Since $n \geq 2$, we see that $m+n \leq 2+\dfrac{2^\ell-1}{2}<2^\ell-1$ when $\ell>2$ (for $\ell=1,2$, we get $2^\ell-1=1,3$ so $n \geq 2 $ is impossible). Thus we can find such a set $S$. If $S=\{s_1, s_2, \ldots, s_k\}$, perform $O(s_1), O(s_2), \ldots, O(s_k)$. I claim that the empty set is not one of $A_1, A_2, \ldots, A_m, B_1, B_2, \ldots, B_n$. Otherwise, based on our operations, we see that originally we must have started off with a set $X=\{s_1, s_2, \ldots, s_k\}=S$. But $S$ was not one of these sets by definition; contradiction! Thus we may indeed assume that the empty set is not one of $A_1, A_2, \ldots, A_m, B_1, B_2, \ldots, B_n$. For each $i$ with $1 \leq i \leq m$, define $f(i), g(i)$ by $A_i=A_{f(i)} \Delta B_{g(i)}$. This is possible because $A_i$ is nonempty, and thus must be representable in this form. Then note that $A_i \Delta B_{g(i)}= (A_{f(i)} \Delta B_{g(i)}) \Delta B_{g(i)} = A_{f(i)} $. Thus $f(f(i))=i$. Furthermore, if $f(i)=i$ then $A_i=A_i \Delta B_{g(i)}$ which implies that $B_{g(i)}=\oslash$; contradiction! Therefore we have a function $f$ from $\{1,2,\ldots, m\}$ into itself with $f(i) \neq i$, $f(f(i))=i$. Thus we can pair off elements of $\{1,2,\ldots, m\}$ into sets $\{i, f(i)\}$. But doing this in succession we can easily see that this implies that $m$ is even. Since $mn=2^\ell-1$ is odd, $m$ is odd; contradiction! therefore our assumption that $n \geq 2$ was wrong. To conclude, the only possibility is $m=2^\ell-1$, $n=1$.