 Polynomial substitution in a size-two Jordan block

For $T=\lambda I+N$ with $N^2=0\ne N$ and a polynomial $p$, the binomial formula gives $p(T)=p(\lambda )I+p^{\prime }(\lambda )N$. Hence its module on the two-dimensional space is indecomposable exactly when $p^{\prime }(\lambda )\ne 0$: in that case the nonzero nilpotent part is one Jordan block; otherwise the operator is scalar and every line is a submodule. For $p(X)=X^2$ over a field of characteristic different from two, the criterion is $\lambda \ne 0$.