$M$ and $S$ have the same characteristic polynomial? [duplicate]

For a rather different approach one can directly prove $p_T(t)=p_{T+N}(t)$ which then gives you the equality of determinants and in turn the equivalence of invertibility. I assume $\dim V = m$ and assume the field has characterisitic zero (or if one wants to contemplate positive characteristic fields, then $\text{char}(\mathbb F)\gt m$)

lemma 1:
by nilpotence, $N^m = \mathbf 0$ and with commutativity this means you can apply the binomial theorem, so $\big(T+N\big)^j= T^j +\Big(\sum_{k=1}^{j-1} \binom{j}{k}T^kN^{j-k}\Big) + N^j$

lemma 2:
$T^k N^r$ for $r\geq 1$ is nilpotent, because
$(T^k N^r)^m = T^{km} N^{rm}= T^{km}\mathbf 0 = \mathbf 0$
and recall that since nilpotent matrices are similar to strictly upper triangular matrices, they necessarily have trace zero

main argument:
$\text{trace}\Big(T+N\Big)= \text{trace}\Big(T\Big)+\text{trace}\Big(N\Big)= \text{trace}\Big(T\big) + 0 $

and for $j\in\big\{1,2,...,m\big\}$
$\text{trace}\Big(\big(T+N\big)^j\Big)= \text{trace}\Big(T^j\Big) + \sum_{k=0}^{j-1} \binom{j}{k}\cdot\text{trace}\Big(T^kN^{j-k}\Big)= \text{trace}\Big(T^j\Big) + 0$

Thus by application of Newton's Identities $\Big(T+N\Big)$ and $\Big(T\Big)$ have the same characteristic polynomial.

Hint: Note that $N$ and $T$ are simultaneously upper triangularizable.