Note $ 1980 = 2^23^25^111^1 $, $ 384=2^7 3^1 $, $ 1694 = 2^1 7^1 11^2 $, and $ 343=7^3 $. Their GCD is $ 1 $, so the integer $ m $ is not a perfect power (i.e., we can't take $ n=1 $). We need $ n=2^a3^b5^c7^d $ (any other prime factors of $ n $ would be superfluous) such that $ (1980+a,384+b,1694+c,343+d) $ has GCD greater than $ 1 $ (i.e., we must use $ n $ to "modify" the exponents of the primes in the prime factorization to get an integer $ mn $ which actually is a perfect power). First we search for a prime which divides at least three of the exponents $ 1980 $, $ 384 $, $ 1694 $, and $ 343 $, which would mean we only have to modify one of them (hence have $ n $ be a prime power). This, however, is only true of the prime $ 2 $, and the exponent not divisible by $ 2 $ is $ 343 $, which is the exponent of 7 in $ m $. Therefore, to modify only one of the exponents, we would need $ (a,b,c,d)=(0,0,0,1) $, giving $ n=7 $. But there is one number less than $ 7 $ which has more than one prime divisor, and that is $ 6 $. Furthermore, $ 7 \mid 1694, 343 $, and $ 1980 \equiv 384 \equiv -1 \mod{7} $, so if we set $ a=b=1 $ and $ c=d=0 $, we find that $ (1980+a,384+b,1694+c,343+d) $ has $ 7 $ as a divisor. This gives $ n=6 $, which is therefore the smallest value such that $ mn $ is a perfect power. In this case, $ mn $ is a perfect $ 7 $th power, so $ k=7 $. Thus $ n+k=6+7=\boxed{13} $.