Let $\mathcal{M}$ denote the set of $d \times d$ positive definite matrices with real entries for some positive integer $d$. Now consider the set
$$ \mathcal{N}=\{A \in M_n(\mathbb{R}): z^TAz > 0, z \in (\mathbb{N} \cup\{0\})^d - \{\mathbb{0}\}\}. $$
Certainly we have $\mathcal{M} \subset \mathcal{N}$, however these sets are not equivalent - for example, when $d=2$, the quadratic form generating $x^2 + y^2 + 3xy \in \mathcal{N}$ but not in $\mathcal{M}$. What additional conditions need to be imposed on quadratic forms to guarantee that they're in the set $\mathcal{N}$?
Of course, $\mathcal{N}$ contains all matrices with $>0$ entries.
On the other hand, changing $A$ into $A+A^T$, the problem reduces to symmetric matrices $A$.
Consider the case $n=2$ (in fact, it was your business) and $A=\begin{pmatrix}a&b\\b&c\end{pmatrix}$. The condition is $ax^2+cy^2+2bxy>0$ when $(x,y)\in \mathbb{N}^2\setminus \{0,0\}$; necessarily $a>0,c>0$; moreover, $at^2+2bt+c>0$ for every $t\in\mathbb{Q}^+$.
That is, $at^2+2bt+c>0$ for every $t\geq 0$ OR
$(*)$ $at^2+2bt+c$ admits a double root in $t_0>0$ a non-rational number.
That implies that, in general, the roots of our polynomial are $<0$ or non-real.
In the first case, the entries of $A$ are $>0$ and, in the second one, $A$ is symmetric $>0$.
An example of the exceptional case $(*)$: $A=\begin{pmatrix}1&-\pi\\-\pi&\pi^2\end{pmatrix}$. Note that $A$ is only symmetric $\geq 0$.
Now, have a look at a generalization, for $n>2$, of the previous result.