Can I diagonalize $\begin{bmatrix} 1 & 1 \\\ 0 & 1 \end{bmatrix}$ over the field of two elements? The minimal polynomial is $(x-1)^2$, but even in $\mathbb{Z}/2\mathbb{Z}$ this can technically factor into $(x+1)(x-1)$ (but aren't 1 and -1 the same)?
Also, can we say that two matrices are similar iff they have the same minimal polynomial?
No matter how you write down the minimal polynomial, the eigenvalue $\lambda = 1 = -1$ appears there twice, so the matrix isn't diagonalizable. You can also compute $PAP^{-1}$ directly for an arbitrary $P$ and see that the resulting matrix cannot be diagonal.
Also, no. Similar matrices have the same minimal polynomial, but not the other way around. See e.g. this answer.
No, because $\begin{bmatrix}1&1\\0&1\end{bmatrix}-I$ has rank one. You can also check by hand that only one of the three non-zero vectors in $\mathbb{F}_2^2$ is an eigenvector.
