Is $a^n b^m$ never regular if n and m have some relation between them?
For a relation $R(n,m)$, let $L_R = \{ a^n b^m : R(n,m) \}$. Call a relation residual if for some $k \geq 1$ and for some $A \subseteq [k]^2$,
$$ R(n,m) \Leftrightarrow (n,m) \pmod{k} \in A. $$
Call a relation semi-residual if it differs from a residual relation by a finite number of points.

Claim: $L_R$ is regular iff $R$ is semi-residual.

Proof: It is not hard to check that if $R$ is semi-residual then $L_R$ is regular. For the other direction, consider an automaton for $L_R$. For each state $\sigma$, let $L_\sigma$ be the set of all $n$ such that upon reading $a^n$, the automaton reaches $\sigma$. Since $L_\sigma$ is unary, it is eventually periodic. By taking the least common multiple of all periods, we see that the state that for $n \geq n_0$, the automaton find itself upon reading $a^n$ depends only on $n\pmod{k_a}$ from some $k_a$. Doing the same for $b^m$, we see that for $n \in L_\sigma$, the set of all $m$ such that $a^n b^m \in L$ is eventually periodic, and so for $m \geq m_0$, depends only on $\sigma$ and on $m\pmod{k_b}$ for some $k_b$. Taking $k$ to be the LCM of $k_a,k_b$, we find that for $n \geq n_0, m \geq m_0$, $a^n b^m \in L$ depends only on $(n,m) \pmod{k}$. Hence the corresponding $R$ is semi-residual.