Yes, a bounded operator can be nonlinear. There are a lot of useful notions of `bounded non-linear operator'. One is that for an operator between topological spaces that the image of compact sets is compact. The operator $Tx = 1/(1-x)$ is bounded on $[0,\infty)$ under this definition, but so are a lot of nasty operators. It depends on what you are trying to get out of your operator.