In the case of the solvable Baumslag-Solitar group $BS(1, n) = \langle a, b \, \vert \, aba^{-1} = b^n \rangle$ with $n \in \mathbb{Z} \setminus \{0\}$, there is only one $T$-system of generating pairs and any number of Nielsen classes, including infinity, can be achieved for a suitable choice of $n$. This [preprint][5] presents of proof of my claim, see Corollary E.
The proof is actually elementary and can be summarized as follows. First identify $G = BS(1, n)$ with $\mathbb{Z}[\frac{1}{n}] \rtimes_n \mathbb{Z}$ where the canonical generator of $\mathbb{Z}$ acts as the multiplication by $n$ on $R = \mathbb{Z}[\frac{1}{n}]$. Then the commutator of a generating pair of $G$ is of the form $(1 - n)u$ for some unit $u$ of $R$ and every unit arises in this way. By Higman's lemma, if two generating pairs $\mathbf{g}_1$ and $\mathbf{g}_2$ of $G$ are Nielsen equivalent then the corresponding units $u_1$ and $u_2$ are related by an element of the form $(-1)^kn^l$ with $k, l \in \mathbb{Z}$. It is not difficult to show that every generating pair of $G$ can be Nielsen reduced to $(a, b')$ for some $b'$ that identifies with a unit of $R$. Now it is immediate to see that if $u_1$ and $u_2$ are related by an element of $(-1)^{\mathbb{Z}}n^{\mathbb{Z}}$ then $\mathbf{g}_1$ and $\mathbf{g}_2$ are Nielsen equivalent. So $G$ has as many Nielsen classes of generating pairs as elements in $R^{\times}/(-1)^{\mathbb{Z}}n^{\mathbb{Z}}$. A careful inspection of $R^{\times}$ shows that the cardinality of the latter set can be infinite (e.g., $n = 6$) or any positive $d \in \mathbb{N}$ (e.g., $n = 2^d$). As the group $G$ has only one $T$-system of generating pairs (see e.g, [3]), it answers OP's question with residually finite instances.
The family of one-relator groups $T(m, n) = \langle a, b \, \vert \, a^m = b^n \rangle$ for $m,n \in \mathbb{Z}$ has been extensively studied and it follows from [1] and [4] (see also [2] for the specific case of the trefoil group) that each $T$-system of generating pairs of $T(m, n)$ (there are infinitely many such $T$-systems if $\vert m \vert,\vert n \vert \ge 2 $ and $\vert m \vert + \vert n \vert > 4$) contains a unique Nielsen equivalence class. This gives a fairly different example from $\mathbb{Z}^2 = BS(1,1)$ for which the number of Nielsen classes in a given $T$-system is obviously $1$.
[1] "Über die gruppen $A^{a}B^{b}=1$", O. Schreier, 1923.
[2] "Presentations of the trefoil group", M. J. Dunwwoody, A. Pietrowsky, 1973.
[3] "Transitivity systems of some one-relator groups", A. M. Brunner, 1974.
[4] "Generators of the free product with amalgamation of two infinite cyclic groups", H. Zieschang, 1977.
[5]: http://arxiv.org/abs/1604.08896