I am not sure what you are asking in your first question (what is a "unit triangulation" ?)
but one can prove the following:

There is no geodesic triangulation of the plane with all vertices of degree >6 and such that
the following conditions holds:

b) the angles of triangles are bounded from below.

Condition b) means that each triangle is bi-Lipshitz to an equilateral triangle, so this is
probably what you ask. And the reason you mention is valid.

Examples of geodesic triangulation with vertices of degree 7 are given in the previous answer,
and it follows from, b) that some triangles in these examples are very thin.

Edited later. My first answer also had condition

a) all triangles are bounded.

I claimed that b) can be replaced with a).

This condition CANNOT be used to replace b).
Here is an example of a geodesic trianglation of the plane with all
vertices of arbitrarily high degree, and bounded triangles.

Begin with an equilateral triange inscribed in the unit circle. From each vertex draw segments
from the unit circle to the concentric circle of radius 2, so that the degree of each vertex
on the unit circle becomes large. Add some chords of the circle of radius 2, connecting the
endpoints of these segmnts. Draw additional segments to make this a triangulation of some
polygon inscribed in the circle of radius 2. All inner vertices of this triangulation
have large degree.

Now repeat this construction. We obtain a sequence of larger and larger polygons $P_n$ each
inscribed in a circle of radius $n$, and each polygon is triangulated so that the
degrees of inner vertices is large. Continuing this indefinitely, we obtain a triangulation of
the plane.