I search for a way to prove that the Jacobson radical of $R=\left [\begin{array}\ \mathbb Z_4 & 2\mathbb Z_4 \\ 0 & \mathbb Z_4 \end{array} \right ]$ is $\left [\begin{array}\ 2\mathbb Z_4 & 2\mathbb Z_4\\ 0 & 2\mathbb Z_4 \end{array} \right ]$, and that this is the set $A(R)= \{a\in R : ann(a)$ is essential in $R\}$.
I could progress by try and error the very tedious job of different cases, but I would thank anybody who tells me how can I prove those facts.
An element $x$ of the ring $R$ is in the Jacobson radical if and only if $1+xr$ is invertible, for each $r\in R$.
Suppose $R=\begin{bmatrix}A & M \\ 0 & A\end{bmatrix}$ is a triangular matrix ring, where $M$ is an $A$-module (with $A$ commutative, for simplicity). The multiplication is $$ \begin{bmatrix}a & m \\ 0 & b\end{bmatrix} \begin{bmatrix}x & z \\ 0 & y\end{bmatrix}= \begin{bmatrix}ax & az+my \\ 0 & by\end{bmatrix} $$ so the first element is invertible if $a$ and $b$ are both invertible. Indeed, $m$ can be arbitrary, because $z=-a^{-1}b^{-1}m$. You can easily check that this right inverse is also a left inverse.
Thus it becomes clear that $$ J(R)=\begin{bmatrix}J(A) & M \\ 0 & J(A)\end{bmatrix} $$ because $$ \begin{bmatrix}1&0\\0&1\end{bmatrix}+ \begin{bmatrix}a & m \\ 0 & b\end{bmatrix} \begin{bmatrix}x & z \\ 0 & y\end{bmatrix}= \begin{bmatrix}1+ax & az+my \\ 0 & 1+by\end{bmatrix} $$ so the invertibility condition reads $a\in J(A)$ and $b\in J(B)$.
The (right) annihilator of $\begin{bmatrix}a & m \\ 0 & b\end{bmatrix}$ is the set of matrices $\begin{bmatrix}x & z \\ 0 & y\end{bmatrix}$ such that \begin{cases} ax=0\\ az+my=0\\ by=0 \end{cases} and verifying essentiality shouldn't be too difficult in your particular case.