Worth trying Vieta's?

$m=1+\sqrt 2\implies (m-1)^2=2\implies \color{#D61F06}{m^2-2m=1}$ Similarly, $\color{#D61F06}{n^2-2n=1}$ .
Given equation can be written as:

$(3(\color{#D61F06}{m^2-2m})-k)(5(\color{#D61F06}{n^2-2n})+k)=16$

$\implies (3(\color{#D61F06}{1})-k)(5(\color{#D61F06}{1})+k)=16$

$\implies k^2+2k+1=0$ $\implies (k+1)^2=0$ $\implies \huge \boxed{k=\color{#3D99F6}{-1}}$

From the given expression, we know that $m+n=2$ and $mn= -1$ , and $m$ and $n$ can be assumed as the roots of the quadratic equation $x^{2}-2x-1=0$ .

Therefore, $m^{2}-2m-1=0, m^{2}-2m=1$ and $n^{2}-2n-1=0, n^{2}-2n=1$ .

Substitute $m^{2}-2m=1$ and $n^{2}-2n=1$ into the given equation, we get $(3(1)-k)(5(1)+k)=16$ .

By solving the equation, our final answer will be $k=\boxed{-1}$ .