Given $E \subseteq \mathbb {F}_q^d \times \mathbb {F}_q^d$, with the finite field $\mathbb {F}_q$ of order q and the integer $d\,\ge \, 2$, we define the two-parameter distance set $\Delta _{d, d}(E)=\{(\|x-y\|, \|z-t\|) : (x, z), (y, t) \in E \}$. Birklbauer and Iosevich [‘A two-parameter finite field Erdős–Falconer distance problem’, Bull. Hellenic Math. Soc. 61 (2017), 21–30] proved that if $|E| \gg q^{{(3d+1)}/{2}}$, then $ |\Delta _{d, d}(E)| = q^2$. For $d=2$, they showed that if $|E| \gg q^{{10}/{3}}$, then $ |\Delta _{2, 2}(E)| \gg q^2$. In this paper, we give extensions and improvements of these results. Given the diagonal polynomial $P(x)=\sum _{i=1}^da_ix_i^s\in \mathbb F_q[x_1,\ldots , x_d]$, the distance induced by P over $\mathbb {F}_q^d$ is $\|x-y\|_s:=P(x-y)$, with the corresponding distance set $\Delta ^s_{d, d}(E)=\{(\|x-y\|_s, \|z-t\|_s) : (x, z), (y, t) \in E \}$. We show that if $|E| \gg q^{{(3d+1)}/{2}}$, then $ |\Delta _{d, d}^s(E)| \gg q^2$. For $d=2$ and the Euclidean distance, we improve the former result over prime fields by showing that $ |\Delta _{2,2}(E)| \gg p^2$ for $|E| \gg p^{{13}/{4}}$.