This question is linked to this one: nilpotent elements of $M_2(\mathbb{R})$, $M_2(\mathbb{Z}/4\mathbb{Z})$
Let $R$ be a commutative ring with unity and let $A\in M_2(R)$. Show that $A$ is a nilpotent matrix iff $\det(A)$ and $\mathrm{trace}(A)$ are nilpotent elements of $R$.
One direction can be found here: Trace of nilpotent matrix over a ring
For the other direction write $A^2=tr(A)A-\det(A)I$ and notice that $tr(A)A$ and $\det(A)I$ are nilpotent and commute.
Another Solution. Claim: Let $A\in M_2(R)$. If $tr(A^{2^k})$ is nilpotent then $tr(A^{2^{k-1}})$ is nilpotent.
By Cayley-Hamilton, $A^{2^k}=(A^{2^{k-1}})^2=tr(A^{2^{k-1}})A^{2^{k-1}}-\det(A^{2^{k-1}})I$.
Thus, $tr(A^{2^k})=tr(A^{2^{k-1}})^2-2\det(A^{2^{k-1}})$. Thus, $tr(A^{2^{k-1}})^2=tr(A^{2^k})+2\det(A^{2^{k-1}})$. Since $tr(A^{2^k})$ is nilpotent, by hypothesis of this claim, and $\det(A)$ is nilpotent then $tr(A^{2^{k-1}})^2$ is nilpotent and $tr(A^{2^{k-1}})$ is too.
Since $A$ is nilpotent exists $k$ such that $tr(A^{2^k})=0$. Now, use the claim $k$ times to obtain $tr(A)$ nilpotent.
Observation: In this answer , it is proved that $tr(A)$ is nilpotent if $A$ is nilpotent. Using the exactly same idea we can prove that the others coefficients of the characteristic polynomial of A (except the first) are also nilpotent.
Now if the coefficients of the characteristic polynomial of $A$ are nilpotent then $A^n$ is a sum of nilpotent matrices that commute: $A^n=\sum_{i=0}^{n-1}a_iA^i$, where $a_i$ are the coefficients of the characteristic polynomial of $A$ (Cayley-Hamilton). Thus, $A^n$ is nilpotent and $A$ is nilpotent.
Thus $A$ is nilpotent iff the coefficients of the characteristic polynomial of $A$ are nilpotent.
I write quickly an answer because this file will be put "on hold". Yes, the ayatollah act even in the mathematical field.
Daniel did a very good job that implies the following equivalence:
Assume that $n!$ is a unit in $R$ and $A\in M_n(R)$. Then $A$ is a nilpotent matrix iff for every $1\leq i\leq n$, $\mathrm{trace}(A^i)$ is a nilpotent element of $R$.
Proof: According to the Qiaochu Yuan'post in Formulas for the (top) coefficients of the characteristic polynomial of a matrix or my post (EDIT 2,3) in nth derivative of determinant wrt matrix the coefficient $a_{n-i}$ of Daniel is $a_{n-i}=\dfrac{1}{i!}(P_{n-i}(trace(A),\cdots, trace(A^{n-i-1}))\pm (n-i-1)trace(A^{n-i}))$ where $P_{n-i}$ is a polynomial with coefficient in $\mathbb{Z}$. According to Daniel's result, the $(a_{n-i})$ are nilpotent ; by a recurrence reasoning, we can conclude that the $(trace(A^k))$ are also nilpotent ; the converse is easy.
Remark: the condition "$n!$ is a unit is essential". Indeed, consider the case $R=\mathbb{Z}/4\mathbb{Z}$ and $n=2$. $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ is nilpotent iff $2|a+d$ and $2|ad-bc$. Now the condition about the traces can be written $2|a+d$ and $2|a^2+d^2+2bc$, that is $2|a^2+d^2$. The entries $b,c$ are free and, consequently, the equivalence does not work.
EDIT. About the orangeskid's answer. Consider the following instance $p(X)=X^3+X+4,R=\mathbb{Z}/15\mathbb{Z}$. Since $p$ admits no roots in $R$, we consider the ring $R_1=R[X]/(p(X))=\{ax^2+bx+c;a,b,c\in R\}$ with the condition $x^3=-x-4$ ; over $R_1$, $p$ admits $3^2=9$ roots ($3$ mod $3$ and $3$ mod $5$):
$x,6x+10,11x+5,6x^2+6x+4,6x^2+x+9,6x^2+11x+14,9x^2+3x+1,9x^2+13x+6,9x^2+8x+11.$
Yet, amongst these roots, we cannot find $\lambda,\mu,\nu$ s.t. $X^3+X+4=(X-\lambda)(X-\mu)(X-\nu)$. Then, as orangeskid wrote, we must consider an additional quotient in order to factorize p into a product of first degree factors.
The traces are more interesting. In dimension $d=2$ we have the implication
$$A^n = 0 \implies (\text{Tr} A)^{2n-1}=0$$ and $2n-1$ is the best exponent.
$\bf{Added}$ Here is a general proof that if $A$ is nilpotent then all the coefficients of the characteristic polynomial of $A$ are nilpotent.
We will use two basic facts from algebra.
$$P(X) = (X-\lambda_1)\cdot \ldots \cdot (X-\lambda_d)$$
Let $A$ a matrix in $M_d(R)$ so that its characteristic polynomial $P_A(X)=\det(X\cdot I_d - A)$ decomposes in the extension $R\subset S$ as
$$P_A(X) = (X-\lambda_1)\cdot \ldots \cdot (X-\lambda_d)$$
Then we have for the characteristic polynomial of $A^n$
$$P_{A^n}(X) = (X-\lambda_1^n)\cdot \ldots \cdot (X-\lambda_d^n)$$
Assume now that $A^n=0$. Then $P_{A^n}(X)=X^d$, that is $$X^d= (X-\lambda_1^n)\cdot \ldots \cdot (X-\lambda_d^n)$$
Plug into this equality $X \mapsto \lambda_i^n$. We get $(\lambda_i^n)^d=0$, that is $\lambda_i^{nd}=0$. From here we conclude that all the fundamental symmetric polynomial in $\lambda_i$ are nilpontent, so the coefficients of the characteristic polynomial of $A$ (other than the leading $1$) are nilpotent, done.
Note that we only used the fact that the coefficients of the characteristic polynomial of $A^n$ are zero. This is weaker than $A^n=0$.
Obs: Using this method we can show that $A^2=0$ implies $(\text{Tr}A)^4$ for $A$ in $M_2(R)$. However, it's true that already implies $(\text{Tr}A)^3=0$. So this method will not give the optimal exponents. Nevertheless, we can have some explicit estimates.
$\bf{Added:}$ Here is an example: Let $A$ a $2\times 2$ matrix in $M_2(S)$ ($S$ commutative ring) and $\lambda_1$, $\lambda_2$ in $S$ so that $$\text{Tr} A = \lambda_1 + \lambda_2 \\ \det A = \lambda_1 \cdot \lambda_2$$ Then we have $$\text{Tr} A^2 = \lambda_1^2 + \lambda_2^2 \\ \det A = \lambda_1^2 \cdot \lambda_2^2 $$
We need to show that $\text{Tr} A^2 = (\text{Tr} A)^2 - 2 \det A$. This could be done directly, or we can use the known identity $A^2 - \text{Tr} A \cdot A + \det A \cdot I_2 = 0$.
