yes, it is the only case. Denote $g(x)=(\int{|f(x,y)|^pd\nu(y))}^{1/p}$, then your condition is $||g||_1=||g||_p$ which is known to imply that $g$ is a constant function, see [here][1].


  [1]: http://en.wikipedia.org/wiki/H%25C3%25B6lder%2527s_inequality