Ideal in the ring of upper triangular matrices

I assume that by "ideal" you mean "two sided ideal".

Note that you can independently scale the two columns or the two rows of a matrix by multiplying on one side or the other by a diagonal matrix. So in a one dimensional ideal the matrices can only have one nonzero entry. There are three locations for this entry and you can check that of the three, only $$\begin{pmatrix} 0 & \ast \\ 0 & 0 \end{pmatrix}$$ gives an ideal, so this is the only possible $1$ dimensional ideal. The ring is three dimensional so all that's left are the two dimensional ideals.

A matrix in $R$ is a unit if and only if it has non-zero entries on the diagonal and we're looking for a proper ideal so all our matrices must have a zero on the diagonal. The ideal must also contain a matrix which has a nonzero on the diagonal (otherwise the ideal is the one dimensional ideal above). We can scale this nonzero entry to $1$ so our ideal must contain a matrix of the form $$\begin{pmatrix} 1 & \ast \\ 0 & 0 \end{pmatrix} \qquad \text{or} \qquad \begin{pmatrix} 0 & \ast \\ 0 & 1 \end{pmatrix}$$ Note it cannot contain one of each because adding them would give a unit. So all the matrices in the ideal must be of the form $$\begin{pmatrix} \ast & \ast \\ 0 & 0 \end{pmatrix} \qquad \text{or} \qquad \begin{pmatrix} 0 & \ast \\ 0 & \ast \end{pmatrix}$$ Now you just have to check that those actually give ideals and then you'll have shown that there are exactly three proper non-trivial ideals.