The matrix $Z^TBZ$ is positive definite and thus nonsingular. We can forget the positive definiteness as long as we keep the nonsingularity in mind. We can also forget the assumption that $A$ has rank $m$ and that $B$ is symmetric.
Let $v \in \ker \left( \begin{array}{ccc}
B & A^T \\
A & 0 \\
\end{array} \right)$. We need to prove that $v = 0$.
Write $v$ in the block-matrix form $v = \left(\begin{matrix} x \\ y \end{matrix}\right)$, where $x$ is a length-$n$ vector and $y$ is a length-$m$ vector. Then, $\left(\begin{matrix} x \\ y \end{matrix}\right) = v \in \ker \left( \begin{array}{ccc}
B & A^T \\
A & 0 \\
\end{array} \right)$, so that $0 = \left( \begin{array}{ccc}
B & A^T \\
A & 0 \\
\end{array} \right) \left(\begin{matrix} x \\ y \end{matrix}\right) = \left(\begin{matrix} Bx+A^Ty \\ Ax \end{matrix}\right)$. Hence, $0 = Bx + A^Ty$ and $0 = Ax$.
We have $0 = Ax$, thus $x \in \ker A$. Hence, $x$ lies in the column space of $Z$ (since the columns of $Z$ span $\ker A$). That is, $x = Zz$ for some vector $z$. Fix such a $z$. We have $0 = B\underbrace{x}_{=Zz} + A^Ty = BZz + A^Ty$, so that $BZz = -A^Ty$. Multiplying this with $Z^T$ from the left, we obtain $Z^TBZz = -\underbrace{Z^TA^T}_{=\left(AZ\right)^T=0 \ \text{(since } AZ = 0 \text{)}}y = 0$. Hence, $z \in \ker \left(Z^TBZ\right)$, so that $z = 0$ (since $Z^TBZ$ is nonsingular).