Since $ \mathbf{M}^2 = \begin{pmatrix} 0 & -5 \\ -2 & 4 \end{pmatrix} \begin{pmatrix} 0 & -5 \\ -2 & 4 \end{pmatrix} = \begin{pmatrix} 10 & -20 \\ -8 & 26 \end{pmatrix}, $ we seek $ p $ and $ q $ such that \[\begin{pmatrix} 10 & -20 \\ -8 & 26 \end{pmatrix} = p \begin{pmatrix} 0 & -5 \\ -2 & 4 \end{pmatrix} + q \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.\]Thus, we want $ p $ and $ q $ to satisfy $ q = 10, $ $ -5p = -20, $ $ -2p = -8, $ and $ 4p + q = 26 $. Solving, we find $ (p,q) = \boxed{(4,10)} $.