Difficulty of forging MACs based on linear functions over $GF\left(2^n\right)$
You are in the right direction.

First, some intuition. The idea is that $y = ax+b$ constitutes a line, and if you need two points in order to fully determine it. However, if you only know one point, and if you don't have any additional information, then it is not possible to determine which line is it since there are "infinitely many" lines passing through the given point (of course, since we're on a finite field there are finitely many, but you get the idea).

Now, for the formalization, imagine that $A$ queries the oracle on some $m$ to get $t = a\cdot m + b$ (let's ignore how $A$ chose $m$, maybe it was crafted in some way, perhaps $m=0$ as you said, but this will turn out to be irrelevant).
We claim that if $A$ successfully crafts another message-tag pair $(m',t')$ with $m=m'$, then it is because $A$ guessed $a$, which is not possible since it was chosen at random from a large enough domain.
Now, this is argued easily by something I already mentioned: given two different points on a line, you can get the line and therefore the coefficients $a,b$.
In other words, once you fix $(m,t)$ with $t = a\cdot m+b$, getting a new $(m',t')$ allows an adversary to find $a$ as $a = (m-m')^{-1}\cdot(t-t')$.
This finishes the argument.