$m=\frac{n-3}{3n-1}\in \mathbb{Z}\Rightarrow \frac{3n-9}{3n-1}=1-\frac{8}{3n-1}\in \mathbb{Z}\Rightarrow 3n-1\in \left\{ \pm 1,\pm 2,\pm 4,\pm 8 \right\}$

3m+3n-9mn=90 (1-3m)(3n-1)=89 By factors that is obvious

$m+n=3mn+30 \Rightarrow 3mn-m-n=-30 \Rightarrow 9mn-3m-3n=-90 \Rightarrow 9mn-3m-3n+1-1=-90$ $\Rightarrow (3m-1)(3n-1)=-89$, which leaves us with 4 cases: (i) $3m-1=-1, 3n-1=89 \Rightarrow m=0, n=30$ (ii) $3m-1=89, 3n-1=-1 \Rightarrow m=30, n=0$ (iii) $3m-1=1, 3n-1=-89$, no integer solutions (iv) $3m-1=-89, 3n-1=1$, no integer solutions Therefore, $(m,n)=(0,30),(30,0)$ are the only integer solutions $\blacksquare$ If only I did this when taking the test...

Can't you use SFFT? $3mn-m-n=-30$, so $mn-m/3-n/3=-10$, so (m-1/3)(n-1/3)=-89/9 then we multiply both sides by 9 i think or something like that to get (9m-3)(9n-3)=-89 wait.. isnt this wrong oops

$m+n=3(mn+10) \iff m-3mn=30-n$ $m(1-3n)=30-n \iff m=\frac{30-n}{1-3n}$ $3m=\frac{90-3n}{1-3n} \iff 3m=\frac{90-3n-1+3n}{1-3n}=\frac{89}{1-3n}$ $3m \in \mathbb{Z} \Rightarrow \frac{89}{1-3n} \in \mathbb{Z}$ $1-3n \in \left\{ \pm 1,\pm 89 \right\}$ if we try this values, we see that: $n \in \left\{ 0, 30 \right\}$ and put them in main equation $(m,n) \in \left\{ (0,30), (30,0) \right\}$ are solutions. We are done.