Suppose I have a matrix:
$\mathbf{Q} = \mathbf{Z}^T\mathbf{B}\mathbf{Z}$
where $\mathbf{Z} \in \mathbb{R}^{n\times m},\mathbf{B}\in \mathbb{R}^{n\times n}, m \leq n$
Can I say that if $\mathbf{Q}$ is positive definite, then $\mathbf{B}$ is also positive definite?
Is this conclusion possibly only valid in certain special cases such as if $\mathbf{Z}$ were a 1D vector, or if $m=n$?
I assume that your matrices are symmetric, as suggested by your title, so that definite positiveness of a matrix $A$ is equivalent to the property that for all nonzero vector $X$, $X^TAX>0$
Simple construction of counter-example: take $B=\begin{pmatrix}1 & 0 & 0 \\ 0& 1& 0\\ 0&0&-1 \end{pmatrix}$ and $Z=\begin{pmatrix}1&0\\0& 1\\0&0\end{pmatrix}$. Then, you have $$Z^TBZ=\begin{pmatrix}1&0&0\\0&1&0\end{pmatrix}\begin{pmatrix}1 & 0 & 0 \\ 0& 1& 0\\ 0&0&-1\end{pmatrix}\begin{pmatrix}1&0\\0& 1\\0&0\end{pmatrix}=\begin{pmatrix}1&0&0\\0&1&0\end{pmatrix}\begin{pmatrix}1&0\\0& 1\\0&0\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}.$$
Changing a bit this example, you can have any counter-example you want for any $n$ and any $m\leq n$.
Actually, your claim true if and only if $Z$ is surjective. Indeed, if $Z$ is surjective, assume that $B$ has a nonpositive eigenvalue $\lambda$ with eigenvector $Y$ and let $X\in \mathbb{R}^m$ such that $ZX=Y$. Then, $X^TQX=\lambda \|Y\|\leq 0$, a contradiction since $X\neq 0$.
Now, if $Z$ is not surjective, it's false. Choose $B$ which is definite positive in the range of $Z$ and definite negative in the orthogonal complement. Then for all $X\neq 0$, $X^TQX>0$ but $B$ is not definite positive.
By the way, note that the opposite claim (if $B$ is definite positive, then so is $Q$) is true if and only if $Z$ is one-to-one (exercise).