Non-nilpotent and non-invertible matrices that have the same characteristic and minimal polynomials have the same Jordan-form

Your argument isn't complete because you haven't used the fact that the vector space is $4$-dimensional. In higher dimensions the statement can be wrong, as the following pair of matrices illustrates: $$ \pmatrix{0&1\\ 0&0\\ &&0&1\\ &&0&0\\ &&&&1}, \ \pmatrix{0&1\\ 0&0\\ &&0&0\\ &&0&0\\ &&&&1}. $$ On $\mathbb C^4$, suppose one of the two endomorphisms --- say $f$ --- has exactly two non-trivial Jordan blocks in its Jordan form. Since $\mathbb C^4$ is $4$-dimensional, the Jordan form of $f$ must be $J_2(\lambda_1)\oplus J_2(\lambda_2)$ for some $\lambda_1,\lambda_2\in\mathbb C$, where the expression $J_k(\lambda)$ denotes a $k\times k$ Jordan block for an eigenvalue $\lambda$. Note that $\lambda_1\ne \lambda_2$, otherwise $f$ is either nilpotent when $\lambda_1=0$, or invertible when $\lambda_1\ne0$. Hence the common minimal polynomial of $f$ and $g$ is $(x-\lambda_1)^2(x-\lambda_2)^2$ and the Jordan form of $g$ is $J_2(\lambda_1)\oplus J_2(\lambda_2)$ too.

The remaining case, where each of $f$ and $g$ has at most one non-trivial Jordan block, should be easy.