There is a "trick", due to Marc Krasner, which prevents you from wasting time in examining "elementary" arithmetic proofs of Fermat's Last Theorem. "Elementary" means precisely that the proof uses only addition and multiplication (operations in a ring), and perhaps also the existence and unicity of decomposition into prime factors (so the ring in question is factorial). I suppose this is the case here, although not all details are given. Then, without checking anything, you can be assured that the reasoning is certainly wrong. This is because all such "elementary" arguments can be repeated word for word in the ring $Z_p$ of p-adic integers, which is factorial (and a lot more !), but in which FLT is false, because in the field $Q_p$ of p-adic numbers, the equation $x^p + y^p = 1$ always has non trivial solutions (if you take $y$ to be a high power of $p$, then p-adic analysis tells you that $1 - y^p$ has a p-th root in $Q_p$).
I think the claim of that thread is blatantly overstated. For one thing, there are lots of properties that $\mathbb{Z}$ has, but $Z_p$ does not have. First and foremost, well ordering of the positive elements, which is heavily used as a key to solving many diophantine equations.
Now consider Fermat's elementary proof that
$$x^4 + y^4 = z^4$$
has no solutions for $(x,y,z) \in \mathbb{Z}$ with $xyz \ne 0$.
I'm not sure whether or not there is a solution in $p$-adic integers, for some prime $p$, but if such solutions exist, it's an example showing that the existence of qualifying $p$-adic solutions doesn't imply the existence of qualifying integer solutions.
As I indicated in the comments, I suspect that most flawed attempts, at least the ones where the solver knows some Number Theory and is not obviously crazy, would include steps for which there is no analogue in $Z_p$, so the $Z_p$ criterion would be useless for invalidating those attempts.
For a given proposed proof, a more common way of quickly demonstrating that there must be a mistake$\,-\,$without actually pinpointing the error, is to observe that the argument would still work for the equation $x^2 + y^2 = z^2$, or, alternatively, to apply the argument line by line for the equation $x^3+y^3=z^3$, and see if the proof at least works for that case.
Best Answer
It's a dumb trick; the author's just misstating the Mason-Stothers theorem, which includes the condition that the three polynomials are relatively prime. Here the polynomials are all multiples of $t$ so they aren't relatively prime.