Proof by induction on $n$, order of matrix.
For $n=1$ we can simply take $Q=1$.
We assume that there exists a common unitary matrix^{} $S$ that triangularizes simultaneously commuting matrices^{} ,$(n1)\times (n1)$.
So we have to show that the statement is valid for commuting matrices, $n\times n$.
From hypothesis^{} $A$ and $B$ are commuting matrices $n\times n$ so these matrices have a common eigenvector^{}.
Let
$Ax=\lambda x$, $Bx=\mu x$ where $x$ be the common eigenvector of unit length and $\lambda $, $\mu $ are the eigenvalues^{} of $A$ and $B$ respectively. Consider the matrix, $R=\left(\begin{array}{cc}\hfill x\hfill & \hfill X\hfill \end{array}\right)$
where $X$ be orthogonal complement^{} of $x$ and ${R}^{H}R=I$, then we have that

$${R}^{H}AR=\left(\begin{array}{cc}\hfill \lambda \hfill & \hfill {x}^{H}AX\hfill \\ \hfill 0\hfill & \hfill {X}^{H}AX\hfill \end{array}\right)$$ 


$${R}^{H}BR=\left(\begin{array}{cc}\hfill \mu \hfill & \hfill {x}^{H}BX\hfill \\ \hfill 0\hfill & \hfill {X}^{H}BX\hfill \end{array}\right)$$ 

It is obvious that the above matrices and also
${X}^{H}BX$, ${X}^{H}AX$ ,$(n1)\times (n1)$ matrices are commuting matrices. Let ${B}_{1}={X}^{H}BX$ and ${A}_{1}={X}^{H}AX$ then
there exists unitary matrix $S$ such that ${S}^{H}{B}_{1}S={\overline{T}}_{2},{S}^{H}{A}_{1}S={\overline{T}}_{1}.$ Now $Q=R\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill S\hfill \end{array}\right)$ is a unitary matrix,
${Q}^{H}Q=I$ and we have

$${Q}^{H}AQ=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill {S}^{H}\hfill \end{array}\right){R}^{H}AR\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill S\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill \lambda \hfill & \hfill {x}^{H}AXS\hfill \\ \hfill 0\hfill & \hfill {\overline{T}}_{1}\hfill \end{array}\right)={T}_{1}.$$ 

Analogously we have that