Define the multi-zeta value $\zeta(p_1,\ldots, p_g)$ as follows: $$ \zeta(p_1,\ldots,p_g) = \sum_{a_1>a_2 >\ldots > a_g\ge 1}\frac{1}{a_1^{p_1}\cdots a_g^{p_g}}, $$ where $p_1\ge 2$ and the other $p_j$ are integers $\ge 1$. Granville and Zagier independently showed that $$ \sum_{p_1+\ldots+p_g= N} \zeta(p_1,\ldots,p_g) = \zeta(N), $$ where in the sum $g$ is fixed, and the sum is over all tuples with $p_1\ge 2$ and other $p_j \ge 1$. This generalizes Euler's relation $$ \zeta(2,1)=\zeta(3). $$ Your result follows from this, upon distinguishing in your harmonic sum when some terms can equal others. For example consider your $m=2$ case which is $$ \sum_{n=1}^{\infty} \frac{1}{n^2} \sum_{m_1\le n} \frac{1}{m_1} \sum_{m_2\le m_1}\frac{1}{m_2} = \sum_{n=1}^{\infty} \frac{1}{n^3} \sum_{m_2 \le n} \frac{1}{m_2} + \sum_{n=1}^{\infty} \frac{1}{n^2} \sum_{m_1 <n} \frac{1}{m_1} \sum_{m_2 \le m_1} \frac{1}{m_2} $$ and the first term is $$ \zeta(4)+\zeta(3,1), $$ while the second term is $$ \zeta(2,2)+\zeta(2,1,1). $$ These add up to $3\zeta(4)$ by Granville-Zagier.

Answer to Question 1.

Define the multi-zeta value $\zeta(p_1,\ldots, p_g)$ as follows: $$ \zeta(p_1,\ldots,p_g) = \sum_{a_1>a_2 >\ldots > a_g\ge 1}\frac{1}{a_1^{p_1}\cdots a_g^{p_g}}, $$ where $p_1\ge 2$ and the other $p_j$ are integers $\ge 1$. Granville and Zagier independently showed that $$ \sum_{p_1+\ldots+p_g= N} \zeta(p_1,\ldots,p_g) = \zeta(N), $$ where in the sum $g$ is fixed, and the sum is over all tuples with $p_1\ge 2$ and other $p_j \ge 1$. This generalizes Euler's relation $$ \zeta(2,1)=\zeta(3). $$ Your result follows from this, upon distinguishing in your harmonic sum when some terms can equal others. For example consider your $m=2$ case which is $$ \sum_{n=1}^{\infty} \frac{1}{n^2} \sum_{m_1\le n} \frac{1}{m_1} \sum_{m_2\le m_1}\frac{1}{m_2} = \sum_{n=1}^{\infty} \frac{1}{n^3} \sum_{m_2 \le n} \frac{1}{m_2} + \sum_{n=1}^{\infty} \frac{1}{n^2} \sum_{m_1 <n} \frac{1}{m_1} \sum_{m_2 \le m_1} \frac{1}{m_2} $$ and the first term is $$ \zeta(4)+\zeta(3,1), $$ while the second term is $$ \zeta(2,2)+\zeta(2,1,1). $$ These add up to $3\zeta(4)$ by Granville-Zagier.