Question: I understand that the dimension of $\mathcal L(V,W)$ is dim$(V)$dim$(W)$ because there is an isomorphism with $\mathbb F^{m,n}$ where $m=$ dim$(V)$ and $n=$ dim$(W)$. This means that there is a basis for $\mathcal L(V,W)$ with dim$(V)$dim$(W)$ elements, right? What's a possible basis?

Answer: 
Once you've chosen a basis $(u\_i)\_{1\le i\le\dim V}$ for $V$ and a basis $(v\_j)\_{1\le j\le\dim W}$ for $W$, a basis for $\mathcal L(V,W)$ is made up of the linear maps $\bigl(\varepsilon\_{ij}\bigr)\_{\substack{1\le i\le\dim V,\\[0.5ex] 1\le j\le\dim W}}$ defined by
$$\begin{cases}\varepsilon\_{ij}(u\_i)=v\_j,\\[0.5ex] \varepsilon\_{ij}(u\_k)=0&\text{if }k\ne i.\end{cases}$$
Yes. And the vector basis are matrices $E\_{i,j}$ such that the $i-j$ th entry is 1 and the others 0:

$$
\begin{pmatrix}
0 & 0 & \cdots & 0 &\cdots & 0\\
0 & 0 & \cdots & 0 &\cdots & 0 \\
\vdots & \vdots &\ddots & \vdots &\ddots & \vdots\\
0 & 0 & \cdots & 1 & \cdots 0\\
\vdots & \vdots &\ddots & \vdots &\ddots & \vdots\\
0 & 0 & \cdots & 0 &\cdots & 0\\
\end{pmatrix}
$$

EDIT

$\mathcal L(V,W)$ is not actually the space if matrices $n\times m$, so they cannot be a basis. My answer should be read in the sense that, a basis for $\mathcal L(V,W)$ are the operators $\varphi\_{i,j}$ such that, given basis $\{v\_i\},\{w\_j\}$ of $V$ and $W$ respectively, their matrix representation are the $E\_{i,j}$.