jmc

Note $1980=2^2{3}^2{5}^{1}11^1$, $384=2^7{3}^1$, $1694=2^1{7}^{1}11^2$, 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 $n=2^a{3}^b{5}^c{7}^d$ (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 $1980\equiv 384\equiv -1\mathrm{mod}7$, 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 $n+k=6+7=\boxed{13}$.