$n^2<m \circ n \to n<23$ So $k \leq 6$ If $k$ is prime then $m=p^{k-1},n=q^{k-1} \to pq|m \circ n-1 \to pq|496 \to pq=2*31 \to$ contradiction If $k=4$ then $m_2m_3=m_4=m,n_2n_3=n_4=n$ and so $m\circ n =(1+n_2m_2)(1+n_3m_3) \to n_2m_2=6,n_3m_3=70 \to $ easy to find contradiction. So $k=6 \to n=p^2q$ and so $n=12,18,20$ $n=12: 1+2m_2+3m_3+4m_4+6m_5+12m_6=497 \to 2|m_3 \to 2|m_2 \to m_2=2,m_3=4$ $1+4+12+4\frac{m}{4}+6\frac{m}{2}+12m=497 \to m=30$ but $4 \not |m \to$ contradiction $n=18: 1+2m_2+3m_3+6m_4+9m_5+18m_6=497 \to 2|m_3 \to m_2=2,m_4=4$ $497=1+4+12+6*\frac{m}{4}+9*\frac{m}{2}+18*m=24m+17 \to m=20$ $18 \circ 20 =1 +2*2+3*4+6*5+9*10+18*20 = 497$ $n=20:1+2m_2+4m_3+5m_4+10m_5+20m_6=497 \to 20m_6 <497-1-2*2-4*3-5*4-10*5 \to 20=n \leq m=m_6<21\to m=20$ But $20 \circ 20 = 546$ Answer $(n,m)=(18,20)$