Key Idea $ $ In the plane $\,\mathbb R^2,\,$ a line $\rm\,a\,x+b\,y = c\,$ of negative slope has points in the first quadrant $\rm\,x,y\ge 0\ $ iff its $\rm\,y$-intercept $\rm\,(0,\,y_0)\,$ is in the first quadrant, i.e. $\,\rm y_0 \ge 0\,.$ We can use an analogous "normalized" point test to check if a discrete line $\rm\,m\,x + n\,y = k\,$ has points in the first quadrant.
Hint $\ $ Normalize $\rm\,k = m\, x + n\, y\,$ so $\rm\,0 \le x < n\,$
by adding a multiple of $\rm\,(-n,m)\,$ to $\rm\,(x,y)$
Lemma $\rm\ \ k = m\ x + n\ y\,$ for some integers $\rm\,x,\,y \ge 0\,$
$\iff$ its normalization has $\rm\,y \ge 0.$
Proof $\ (\Rightarrow)\ $ If $\rm\ x,\, y \ge 0\,$ normalizing adds $\rm\,(-n,m)\,$ zero or more times, preserving $\rm\,y \ge 0\,.\,$
$(\Leftarrow)\ \,$ If the normalized rep has $\rm\ y < 0,\,$ then $\rm\,k\,$ has no representation with $\rm\, x,\,y \ge 0\,\, $ since to shift so that $\rm\,y > 0\,$
we must add $\rm\,(-n,m)\,$ at least once, which forces $\rm\,x < 0,\,$ by $\rm\,0\le x < n.\ \ \ $ QED
To solve the OP, note that since $\rm\, k = m\ x + n\ y\, $ is increasing in both $\rm\,x\,$ and $\rm\,y,\,$ it is clear that the largest non-representable $\rm\,k\,$ has normalization $\rm\,(x,y) = (n\!-\!1,-1),\,$ i.e. the lattice point that is rightmost (max $\rm\,x$) and topmost (max $\rm\,y$) in the nonrepresentable lower half $\rm (y < 0)$ of the normalized strip, i.e. the vertical strip where $\rm\, 0\le x \le n-1.\,$ Thus $\rm\,(x,y) = (\color{#0a0}{n\!-\!1},\color{#c00}{-1})\,$ yields that $\rm\, k = mx+ny = m\,(\color{#0a0}{n\!-\!1})+n\,(\color{#c00}{-1}) = mn\! -\! m\! -\! n\ $ is the largest nonrepresentable integer. $\ $ QED
Notice that the proof has a vivid geometric picture:
representations of $\rm\,k\,$ correspond to lattice points $\rm\,(x,y)\,$
on the line $\rm\, n\ y + m\ x = k\,$ with negative slope $\rm\, = -m/n\,.\,$
Normalization is achieved by shifting forward/backward
along the line by integral multiples of the vector $\rm\,(-n,m)\,$
until one lands in the normal strip where $\rm\,0 \le x \le n-1.\,$ If the normalized rep has $\rm\,y\ge 0\,$ then we are done; otherwise, by the lemma, $\rm\,k\,$ has no rep with both $\rm\,x,y\ge 0\,.\,$ This result may be viewed as a discrete analog of the fact that, in the plane $\,\mathbb R^2,\,$ a line of negative slope has points in the first quadrant $\rm\,x,y\ge 0\ $ iff its $\rm\,y$-intercept $\rm\,(0,\,y_0)\,$ lies in the first quadrant, i.e. $\rm y_0 \ge 0\,.$
There is much literature on this classical problem. To locate such work
you should ensure that you search on the many aliases,
e.g. postage stamp problem, Sylvester/Frobenius problem,
Diophantine problem of Frobenius, Frobenius conductor,
money changing, coin changing, change making problems,
h-basis and asymptotic bases in additive number theory,
integer programming algorithms and Gomory cuts,
knapsack problems and greedy algorithms, etc.