jmc

Let $p$ and $q$ be two primes that are roots of $x^2-63x+k=0$. Then $$x^2-63x+k=(x-p)(x-q)=x^2-(p+q)x+p\cdot q,$$ so $p+q=63$ and $p\cdot q=k$. Since $63$ is odd, one of the primes must be $2$ and the other $61$. Thus, there is exactly $\boxed{1}$ possible value for $k$, namely $k=p\cdot q=2\cdot 61=122$.