Prove that $A$ and $B$ are nilpotent - proof checking 
Let $A$ and $B$ be $n \times n$ matrices with real entries and $c_1, c_2, \dots ,c_{n+1}$ distinct real numbers such that $A+c_1B, A+c_2B \dots, A+c_{n+1}B$ are nilpotent matrices. Prove that $A$ and $B$ are nilpotent. 

Is my proof alright? This is my approach: 
For $k$ chosen arbitrarily from $\{1,2, \dots, n\}$ define the polynomial :
$ P_k(x) = \operatorname{tr}[ (A+xB)^k]$.
Suppose that $P_k$ is not constant taking the value $0$.
$A+c_iB \text{ is nilpotent} \iff tr[ (A+c_iB)^k ] = 0 \forall k \in \mathbb{N}^* \text{ and } \forall i \in \{1,2,\dots, n+1\} \Rightarrow P_k(c_1) = P_k(c_2) = \cdots = P_k(c_{n+1}) = 0 \Rightarrow$
$$\operatorname{deg} P_k \geq n+1.$$ 
However the elements of $(A+xB)^k$ are polynomials in $x$ of degree at most $k$. So $$ \operatorname{deg} P_k \leq k. $$
But from these two statements $k \geq n+1$ which is a contradiction because we chose $ k \leq n$. So $P_k(x) = 0 \forall x \in \mathbb{R} $. That means that all of its coefficients are $0$. But, because $(A+xB)^k = x^k B^k + \dots + A^k$, the leading coefficient of $P_k$ is $\operatorname{tr} (B^k)$ and the constant term is $\operatorname{tr} (A^k) \Rightarrow \operatorname{tr} (A^k) = \operatorname{tr} (B^k) = 0 \forall k \in \mathbb\{1,2 \dots n\} \Rightarrow $ $$ A \text{ and } B \text{ are nilpotent} $$
 A: Consider $(A+xB)^n$. This is a matrix where each element is a polynomial of degree at most $n$ in $x$. Furthermore, each of these polynomials is zero for $x=c_i, 1\le i \le n+1.$ 
Therefore each element is zero for all $x$. The value $x=0$ shows that $A$ is nilpotent. 
Then $(B+xA)^n$ is zero for all non-zero $x$ and therefore zero for all $x$. Hence $B$ is also nilpotent.
A: Your proof is correct, but I would say that it is a bit difficult to follow at some points. Here's an edit of your proof that I would find easier to read.
Note that a matrix $M$ is nilpotent if and only if we have $\operatorname{tr}(M^k) = 0$ for $k = 1,\dots,n$. With that in mind: for $k = 1,2,\dots,n,$ define the polynomial $P_k(x) = \operatorname{tr}[ (A+xB)^k]$. Fix any such $k$.
Suppose that $P_k$ is not identitcally $0$. We note that because $A + c_i B$ is nilpotent for $i = 1,\dots,n+1$, we must have $P_k(c_i) = 0$ for all such $i$.  This means that $P_k(x)$ is a polynomial with degree $k \leq n$ with $n+1$ zeros.  However, if $P_k$ is not identitcally zero, then this is impossible.
So $P_k(x)$ is identitcally zero. That means that all of its coefficients are $0$. But, because $(A+xB)^k = x^k B^k + \dots + A^k$, the leading coefficient of $P_k$ is $\operatorname{tr} (B^k)$ and the constant term is $\operatorname{tr} (A^k).$ So, $\operatorname{tr} (A^k) = \operatorname{tr} (B^k) = 0$ for $k = 1,\dots,n$.
This implies that $A$ and $B$ are nilpotent, which was the desired conclusion.
