A rhombus with diagonals $\sqrt{21}$ and $\sqrt{31}$ has side $\sqrt{\frac{21}{4} + \frac{31}{4}} = \sqrt{13},$ so the three edge vectors $\mathbf{u},$ $\mathbf{v},$ $\mathbf{w}$ all have squared length $13.$ In the face spanned by $\mathbf{u}$ and $\mathbf{v}$ the diagonals are $\mathbf{u} \pm \mathbf{v},$ with $|\mathbf{u} \pm \mathbf{v}|^2 = 26 \pm 2\,\mathbf{u}\cdot\mathbf{v};$ matching $\{21, 31\}$ gives $\mathbf{u}\cdot\mathbf{v} = \pm\frac{5}{2},$ and likewise for the other two pairs.

The squared volume is the Gram determinant $$\begin{aligned} V^2 &= \det\begin{pmatrix} 13 & x & y \\ x & 13 & z \\ y & z & 13 \end{pmatrix} \\ &= 2197 - 13(x^2 + y^2 + z^2) \\ &\quad {}+ 2xyz \\ &= 2197 - \frac{975}{4} \\ &\quad {}+ 2xyz \end{aligned}$$ with $x, y, z \in \left\{\pm\frac{5}{2}\right\}.$ Negating an edge vector flips the signs of two of $x, y, z,$ so only the sign of $xyz$ matters: $2xyz = \pm\frac{125}{4},$ giving $V^2 = \frac{7938}{4}$ or $\frac{7688}{4}.$

The ratio of the volumes is $\sqrt{\frac{7938}{7688}} = \sqrt{\frac{3969}{3844}} = \frac{63}{62},$ already in lowest terms, so $m + n = 63 + 62 = 125.$