I use boldface (e.g. $\mathbf{C},\mathbf{D}$) for categories and fraktur (e.g. $\mathfrak{B}$ for bicategories.)
According to this blog post at nLab, with some of the notation changed a little:
If $1$ denotes the terminal bicategory, then a lax functor $\mathfrak{B} \leftarrow 1$ is the same as a monad in $\mathfrak{B}$.
Sounds cool! Unfortunately, I don't get it.
As I understand it, a monad in $\mathfrak{B}$ is an object $\mathbf{C}$ of $\mathfrak{B}$ (which is a $0$-cell), together with an endomorphism $F$ of $\mathbf{C}$ (which is a $1$-cell), together with morphisms $\eta : F \leftarrow \mathrm{id}_\mathbf{C}$ and $\mu : F \leftarrow F \circ F$ (which are $2$-cells) satisfying certain identities.
Now write $1$ for the terminal bicategory, $1_0$ for its sole object, $1_1$ for its sole morphism, and $1_2$ for its sole $2$-cell.
Let $\varphi : \mathfrak{B} \leftarrow 1$ denote a lax functor.
Then $\varphi(1_0)$ is an object of $\mathfrak{B}$. So define $\mathbf{C} = \varphi(1_0)$.
And $\varphi(1_1)$ is an endomorphism of $\mathbf{C}$. So define $F = \varphi(1_1)$.
And $\varphi(1_2)$ is... an endomorphism of $F$? Umm, what? Where do $\mu$ and $\eta$ come from? And where do the equations between $\mu$ and $\eta$ come from?
Because $1_1$ is the identity, there is a natural isomorphism $1_1 \circ 1_1 \cong 1_1$. The fact that $\varphi$ is a lax functor turns this into a natural transformation $F \circ F \to F$ which is not required to be a natural isomorphism; this is what it means to laxly preserve composition. Similarly, what it means to laxly preserve identities is that instead of $F$ being naturally isomorphic to the identity (which is what a real functor would do), there is a natural transformation $\text{id}_C \to F$ which is not required to be a natural isomorphism.
The rest of the lax functor axioms become the monad axioms; you can use this to guess what they are.