Why is $\det(I + A^{50}) = 4$ here?

Let $D$ be the diagonal matrix in question.

Then, $I+A^{50}=I+TD^{50}T^{-1}=T(I+D^{50})T^{-1}$ and hence,

$$ \det(I+A^{50})=\det(T)\det(I+D^{50})\det(T^{-1})=\det(I+D^{50}) $$ Can you take it from here?

As you said, you can diagonalize $A=PDP^{-1}$ with $D=\operatorname{diag}(1,-1,0)$. Then $A^{50}=PD^{50}P^{-1}$. $D^{50}=\operatorname{diag}(1,1,0)$. So, $$\det(I+A^{50})=\det(P^{-1}(I+A^{50})P)=\det(I+D^{50})=\det(\operatorname{diag}(2,2,1))=4.$$ The first equality follows from the fact that determinants are invariant under change of basis. The second follows from distributing.

If $v_1$, $v_{-1}$, $v_0$ are respective eigenvectirs to eigenvalues $1,-1,0$, then they are necessarily linearly independent and, by dimension, form a basis. We have $(I+A^{50})v_\lambda=v_\lambda+\lambda^{50}v_\lambda$, so $v_1\mapsto 2v_1$, $v_{-1}\mapsto 2v_{-1}$, $v_0\mapsto v_0$. From this, clearly $\det(I+A^{50})=2\cdot 2\cdot 1=4$.