We know that:
$$\,_2F_1(a,b,b,z)=(1-z)^{-a}$$
But putting $b<0$ with $b$ integer Mathematica does not use the previous formula but generates the hypergeometric polynomial. For instance:
$$\,_2F_1(a,-2,-2,z)= \frac{1}{2} (2 + 2 a z + a z^2 + a^2 z^2)$$
I don't understand why this happens because the 2 pochammer should cancel in the hypergeometric definition also if they are negative. Any Idea?
This comes down to how Mathematica interprets these types of special cases for the Hypergeometric function. Specifically, it is using this definition so that your hypergeometric function is interpreted as $$ \begin{aligned} F(a,-2;-2;z) &=\lim_{c\to-2}\lim_{b\to-2}F(a,b;c;z)\\ &=\lim_{c\to-2}\frac{a^2 z^2-2 a c z+a z^2-2 a z+c^2+c}{c (c+1)}\\ &=\frac{1}{2} (2 + 2 a z + a z^2 + a^2 z^2)\\ &=1 +a z +\frac{a(1+a)z^2}{2}\\ &=\sum_{k=0}^2 \frac{(a)_k}{k!}z^k. \end{aligned} $$
Here the Mathematica code:
Limit[Hypergeometric2F1[a, -2, c, z], c -> -2].So we see that the $(-2)_k$ terms do cancel out; however, the series also truncates to only the first three terms and thus does not gives the binomial result you expected.