I start with definitions.
Definition 1. A linear space is a pair $(X,\mathcal L)$ consisting of a set $X$ and a family $\mathcal L$ of subsets of $X$ satisfying three axioms:
(L1) for any distinct points $x,y\in X$ there exists a unique line $L\in\mathcal L$ containing $x$ and $y$;
(L2) every set $L\in\mathcal L$ contains at least three points;
(L3) $X\notin\mathcal L$.
Elements of the family $\mathcal L$ are called lines.
For any distinct points $x,y\in X$ of a linear space $(X,\mathcal L)$, the unique line $L\in\mathcal L$ containing the points $x,y$ will be denoted by $\overline{xy}$.
Definition 2. A linear space $(X,\mathcal L)$ is called
$\ \bullet$ uniform if all lines in $X$ have the same cardinality;
$\ \bullet$ hyperbolic if for every points $\ x,o,y\in X$ and $p\in \overline{xy}\setminus(\overline{ox}\cup\overline{oy})\,\ $ the set
$\ \{u\in\overline{oy}\,:\ \overline{up}\ \cap\overline{ox}\ = \varnothing\}\ $ contains more than one point.
Question. Is every uniform hyperbolic linear space infinite?
Added in Edit, after answers of @ihromant: It turns out that typical examples of finite uniform hyperbolic linear spaces (including those two in the second answer of @ihromant) are provided by classical unitals in finite projective planes of square order $q^2$ for some $q$ which is a power of a prime number. By Lemma 7.42 in the book "Unitals in projective planes", every classical unital $U$ in a projective plane $PG(2,q^2)$ contains no Pasch configurations, which means that for every points $o,x,y\in U$, $p\in\overline{oy}\setminus(\overline{ox}\cup\overline{oy})$, and $u\in\overline{oy}\setminus\{o,y\}$ the lines $\overline{up}$ and $\overline{ox}$ are disjoint. So $U$ is hyperbolic in a very strong sense. It is a longstanding conjecture that the absence of Pasch configurations characterizes classical unitals, see page 161 of the mentioned book "Unitals in projective planes". This conjecture is confirmed by Wilbrink under some additional conditions on the geometry of a unital, see Theorem 7.43, 7.44 in the book. I recall that a linear space $(X,\mathcal L)$ is a unital if $|X|=n^3+1$ for some number $n$ such that all lines $L\in\mathcal L$ have cardinality $n+1$. A classical unital in a projective plane $PG(2,q^2)$ is the subset defined by the equation $X^{q+1}+Y^{q+1}+Z^{q+1}=0$ in the homogeneous coordinates. Classical unitals have many exciting geometric properties, discussed in the book. The geometry of the classical unital in the projective plane $PG(2,9)$ is visualized in this MO-post.
Best Answer
The answer is "No".There exist a finite plane with required property.
First of all, let's rephrase your question. Your uniform linear space is a synonym to a BIBD (balanced incomplete block design) with λ = 1. There exists an example of a BIBD(217, 7, 1): v = 217, k = 7 so-called "Difference family" where blocks (in your case lines) are obtained by cyclic shifts by addition of some base sets.
So, there are 6 base sets for your family (please note that the last one produces only 31 lines):
This plane consists of 217 points, 217 * 5 + 31 = 1116 lines, there are 36 lines that go through every point. Through every point that does not lie on a given line you have 29 "parallel" lines. And last: hyperbolic index (number of non-crossing lines from your definition) is 2..5. This properties are checked by Java code present here: