Note, a polyhedron is the intersection of finitely many half spaces in $\mathbb{R}^n$ and a polytope is a bounded polyhedron.
Preliminary: Consider a hyperplane $C_1x_1 + C_2x_2 + \dots + C_nx_n + D = 0$ not passing through the origin. Since not passing through the origin, the term $D \ne 0$ and then both equation sides can be divided by $(-D)$. Therefore, a hyperplane not passing through the origin can be simply represented by $a_1x_1 + a_2x_2 + \dots + a_nx_n =1$ where $a_i = -C_i / D$. A hyperplane always cuts the whole space into two half-spaces, and this simple equation ensures that any point in the same half-space as the origin always makes the inequality $a_1x_1 + a_2x_2 + \dots + a_nx_n <= 1$ hold.
Question: Consider a polyhedron $S$ (may be unbounded) formed by $m$ hyper planes not passing through the origin. We assume that $m>n$ throughout the question since $S$ cannot be bounded when $m \le n$.
$$S := \{ \mathbf{x} \in \mathbb{R}^n: \mathbf{A}\mathbf{x} <= \mathbf{b}\},$$
where
$$\mathbf{A} := [a_{11}, a_{12}, \dots , a_{1n}; \\ a_{21}, a_{22}, \dots , a_{2n}; \\ \dots ; \\a_{m1}, a_{m2}, \dots , a_{mn};] $$
is a $m$ by $n$ matrix and $\mathbf{b} := [1;1;...;1] \in \mathbb{R}^m$.
I am wondering which constraints on the coefficients $\mathbf{A}$ can guarantee that the polyhedron $S$ is a bounded polytope? (Of course $S$ is nonempty. The origin $ \in S$ because $a_{k1}0 + a_{k2}0 + \dots +a_{kn}0 = 0 <= 1, k \in \{1,...,m\}$).
Lemma: $S$ is unbounded iff there is some $x\neq 0$ satisfying $Ax \le 0$.
Suppose $S$ is unbounded, then there are $x_n \in S$ such that $\|x_n\| \to \infty$, and $A {x_n \over \|x_n\|} \le {b \over \|x_n\|}$. A compactness/continuity argument shows that some unit $x$ exists that satisfies $Ax \le 0$. For the other direction, suppose $x\neq 0$ satisfies $Ax \le 0$, then clearly $nx \in S$ for all $n$.
Note that the existence of a non zero $x$ satisfying $Ax \le 0$ is equivalent to the existence of a non zero $x$ satisfying $Ax \ge 0$, so to avoid confusing myself I will address the latter.
Let $Q= \{ x | x \ge 0 \}$ be the positive quadrant and $\Sigma = \{ x \in Q| \sum_k x_k = 1 \}$.
We are looking for a non zero $x \in {\cal R}A \cap Q$. Since both ${\cal R}A, Q$ are both cones, it is equivalent to determining if ${\cal R}A $ intersects $\Sigma$.
Note that if $\ker A$ is non trivial, then $S$ is clearly unbounded, so I will assume that $\ker A$ is trivial subsequently.
The problem reduces to solving $\min_{\mu \in \Sigma} \min_x \|Ax-\mu\|$. If we solve the inner least squares problem and let $B= A (A^T A)^{-1} A^T - I$, the problem becomes $\min_{\mu \in \Sigma} \|B\mu\|$.
The last problem is equivalent to $\min_{y \in C} \|y\|$, where $C = \operatorname{co} \{ B e_k \}$.
Hence $S$ is bounded iff $\ker A$ is non trivial and $\min_{y \in C} \|y\| > 0$.