You cannot write this system in the form you want. The wikipedia equations you are citing and trying to get are linear. The system you have is a nonlinear Lagrangian system. There is a conceptually very deep method that can put your system as an explicit first order system (i.e. involving only first derivatives) of four functions and four differential equations. It is called Legendre's transformation which transforms the Euler-Lagrange equations into Hamilton's equations.
Start with the Lagrangian function
$$\mathcal{L}(z, \theta, \dot{z}, \dot{\theta}) = \frac{M+m}{2} \, \dot{z}^2 + mL\,\cos{\theta}\,\, \dot{z}\,\dot{\theta} + \frac{mL^2}{2} \, \dot{\theta}^2 + u\, z - mgL \, \cos{\theta}$$ Compute
$$\frac{\partial \mathcal{L}}{\partial \dot{z}} = (M+m) \, \dot{z} + mL \,\dot{\theta} \cos{\theta} $$
$$\frac{\partial \mathcal{L}}{\partial \dot{\theta}} = mL \, \dot{z} \cos{\theta + mL^2\, \dot{\theta}}$$
$$\frac{\partial \mathcal{L}}{\partial {x}} = u$$
$$\frac{\partial \mathcal{L}}{\partial {\theta}} = - \, mL \, \dot{z}\, \dot{\theta} \sin{\theta} + mgL \, \sin{\theta}$$
The Euler-Legrange equations are
\begin{align}
\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{z}}\right) &= \frac{\partial \mathcal{L}}{\partial {z}} \\
\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{\theta}} \right) &= \frac{\partial \mathcal{L}}{\partial {\theta}}
\end{align} or written explicitly
\begin{align}
&\frac{d}{dt}\left((M+m) \, \dot{z} + mL \,\dot{\theta} \cos{\theta}\right) = u \\
&\frac{d}{dt}\left(mL \, \dot{z} \cos{\theta + mL^2\, \dot{\theta}}\right) = - \, mL \, \dot{z}\, \dot{\theta} \sin{\theta} + mgL \, \sin{\theta}
\end{align} which after differentiation in the left hand side become
\begin{align}
&(M+m) \, \ddot{z} - mL \,\dot{\theta}^2 \sin{\theta} + mL \,\ddot{\theta} \cos{\theta} = u \\
&mL \, \ddot{z} \cos{\theta} - mL \,\dot{z} \, \dot{\theta} \sin{\theta} + mL^2\, \ddot{\theta}= - \, m L \, \dot{z}\, \dot{\theta} \sin{\theta} + mgL \, \sin{\theta}
\end{align}
They reduce to
\begin{align}
&(M+m) \, \ddot{z} - mL \,\dot{\theta}^2 \sin{\theta} + mL \,\ddot{\theta} \cos{\theta} = u \\
&m \, \ddot{z} \cos{\theta} + mL\, \ddot{\theta} = mg \, \sin{\theta}
\end{align}
With this structure in mind we switch to Hamiltonian formulation. Define the conjugate momenta
\begin{align}
p_z = \frac{\partial \mathcal{L}}{\partial \dot{z}} &= (M+m) \, \dot{z} + mL \,\dot{\theta} \cos{\theta}\\
p_{\theta} = \frac{\partial \mathcal{L}}{\partial \dot{\theta}} &= mL \, \dot{z} \cos{\theta + mL^2\, \dot{\theta}}
\end{align} Solve this linear system with respect to $\dot{z}, \dot{\theta}$ obtaining
\begin{align}
\dot{z} &= \frac{mL^2 \, p_{z} - mL \, p_{\theta} \cos{\theta}}{(M+m)mL^2 - m^2L^2 \, \cos^2{\theta}} = \frac{mL^2 \, p_{z} - mL \, p_{\theta} \cos{\theta}}{MmL^2 + m^2L^2 \, \sin^2{\theta}} \\
\dot{\theta} &= \frac{- mL \, p_{z} \cos{\theta} + (M+m) \, p_{\theta}}{(M+m)mL^2 - m^2L^2 \, \cos^2{\theta}} = \frac{- mL \, p_{z} \cos{\theta} + (M+m) \, p_{\theta}}{MmL^2 + m^2L^2 \, \sin^2{\theta}}
\end{align}
Now observing that
\begin{align}
\frac{d}{dt}p_z &= \frac{\partial \mathcal{L}}{\partial {z}} \\
\frac{d}{dt} p_{\theta} &= \frac{\partial \mathcal{L}}{\partial {\theta}}
\end{align} combining all pairs of equations together
\begin{align}
\dot{x} &= \frac{L \, p_{z} - \, p_{\theta} \cos{\theta}}{ML + mL \, \sin^2{\theta}} \\
\dot{\theta} &= \frac{- mL \, p_{z} \cos{\theta} + (M+m) \, p_{\theta}}{MmL^2 + m^2L^2 \, \sin^2{\theta}}\\
\dot{p}_x &= u \\
\dot{p}_{\theta} &= - \, \, \frac{\big(L \, p_{z} - \, p_{\theta} \cos{\theta}\big) \, \big(- mL \, p_{z} \cos{\theta} + (M+m) \, p_{\theta}\big) \, \sin{\theta}}{\big(ML + mL \, \sin^2{\theta}\big)^2} \, + \, mgL \, \sin{\theta}
\end{align} These are the Hamiltonian equations.
I am assuming the function $u$ is some kind of control function.
