Trigonalization and eigenvalues

Forward direction. If $T=P^{-1}AP$ is upper triangular, then $\chi_A=\chi_T=\prod_i(X-t_{ii})$, which splits over $\mathbb{R}$.

Converse, by induction on $n$. For $n=1$ every matrix is triangular.

Let $n\ge2$ and assume $\chi_A$ splits over $\mathbb{R}$. Then $\chi_A$ has a real root $\lambda_1$, hence an eigenvector $v_1\neq0$ with $Av_1=\lambda_1v_1$. Complete $(v_1)$ to a basis of $\mathbb{R}^n$ and let $P_1$ be its matrix. Then
$$M=P_1^{-1}AP_1=\begin{pmatrix}\lambda_1&c\\0&B\end{pmatrix}$$
with $B$ an $(n-1)\times(n-1)$ matrix and $c$ a row vector. We have $\chi_A=\chi_M=(X-\lambda_1)\chi_B$, so $\chi_B$ divides a polynomial that splits over $\mathbb{R}$, hence $\chi_B$ splits over $\mathbb{R}$ (unique factorization).

By the induction hypothesis there is $Q\in GL_{n-1}(\mathbb{R})$ with $T=Q^{-1}BQ$ upper triangular. Let $R=\begin{pmatrix}1&0\\0&Q\end{pmatrix}$. Then
$$R^{-1}MR=\begin{pmatrix}\lambda_1&cQ\\0&T\end{pmatrix},$$
which is upper triangular. So $A$ is trigonalizable, with $P=P_1R$.

Conclusion. $A$ is trigonalizable over $\mathbb{R}$ if and only if all roots of $\chi_A$ are real. In particular every complex matrix is trigonalizable over $\mathbb{C}$.