Do a central reflection of your triangle in the midpoint of any one of its sides to get a new triangle that shares that side, but is otherwise disjoint from the original triangle; the two together form a parallelogram. One easily sees that the hypotheses about the triangle imply that parallelogram still has no interior lattice points, but its four vertices are lattice points.
To show intuitively that the parallelogram must have area $1$, pave the whole plane by translates of the parallelogram laid side-to-side in the obvious way. Now each lattice point is the bottom left corner (for some appropriate meaning of "bottom-left") of a unique parallelogram. Choosing a large convex region, so that the number of parallelograms crossing its boundary is small with respect to the number completely contained in it, the latter number is approximately equal to the number of lattice points in the region, which is approximately equal to the area of the region. Taking the limit as the size of the region goes to infinity, one sees that the area of individual parallelograms must be$~1$.
A somewhat more formal way to complete that argument is to roll up the plane to a torus: take the projection $\def\R{\Bbb R}\def\Z{\Bbb Z}\R^2\to\R^2/(n\Z\times m\Z)$ for some positive integers $n,m$. The codomain is a compact surface of area $nm$, which contains $nm$ points that are the image of a lattice points (element of $\Z\times\Z$). Choosing preimages for those points, and taking the corresponding set of parallelograms, one obtains a region that under the projection covers our compact surface exactly once; since the projection is locally area-preserving, that region has area $nm$, and each parallelogram has area$~1$. In fact one could just choose $n=m=1$.
First, we simplify the problem by allowing for signed distances, and consider points outside the triangle.
We will show that the set of points that satisfy $ax+by+cz=k$ is either the null set, a line or the entire plane. Then, when restricted to the (interior) of the triangle, this becomes a null set, a line segment, or the entire interior, so the result follows.
Hint: For a point $P$, if the side lengths of the triangle are $l_a, l_b, l_c$, then $l_a x + l_b y + l_c z = 2 \Delta $.
Hint: Hence, it follows that $ax+by+cz=k$ is the entire plane iff $a:b:c:k = l_a : l_b: l_c : 2 \Delta$.
If $a:b:c = l_a : l_b: l_c $, then we either get the entire plane or the null set.
Henceforth, assume $a:b:c \neq l_a : l_b: l_c $.
Hint: For fixed values of $a, b, c, k$, if $P_1$ and $P_2$ are distinct points that satisfy the equation, then so does the entire line segment $P_1 P_2$.
Corollary: The solution set is a (affine) sub-space. It remains to show that the solution set cannot be a single point.
Hint: For fixed values of $a, b, c, k$, if there is (at least) 1 point that satisfies it, then there are at least 2 distinct points that satisfy it.
Suppose we have the solution $P_1 = (x_1, y_1, z_1)$, then the point $P_2 = (x_1 + bl_c - cl_b, y_1 +cl_a - al_c, z_1 + al_b-bl_a)$ will also satisfy both equations (expand and cancel terms). These are distinct points if $a:b:c \neq l_a : l_b: l_c $.
Best Answer
One other idea besides scan-line:
I still think a well-implemented scan-line algorithm would be faster because it does not have to apply the point-test for every point, but only has to find the start and end of a scan-line. This can be done fast if you translate the triangle into a half-space representation, i.e. a point is an inner point if it satisfies
$$a_ix +b_iy <1,\quad i=1,2,3$$
for appropriate $a_i,b_i,i=1,2,3$ which you need to compute only once.