Decide if a specific Turing machine halts on a specific string
The problem you are referring to is sometimes known as the Acceptance problem. It is indeed undecidable.

Let me attempt to find the source of your confusion. I assume by linking the other thread you are referring to the following decidable language:

$$
L_{HP-M'-x} = 
\begin{cases}
1 &amp;  \text{M' halts on x} \\
0 &amp;  \text{otherwise}
\end{cases}
$$

The reason this language is decidable is because, given any $M'$ and $x$, this language must be either $\{1\}$ or $\{0\}$. In particular, this MUST be a finite language, and ALL finite languages are decidable.

Note that this doesn't mean we can solve the Acceptance problem. Using classical logic, we deduce that either a machine accepts a string or it does not. The hard part is figuring out which case it is! We know that the language is decidable, but we do not know what a machine that accepts this language looks like because we don't know what the language itself looks like.