Your question is somewhat ill-posed: given $n$ and $k$, it is easy to construct a Tanner graph with a 4-cycle and column weight at most 2.

Instead, you can ask the question of what is the maximum column weight of a Tanner graph, given that is has no 4-cycle. Assuming uniform column weights, then the maximum column weight is $O(\sqrt{n})$. This follows from a result in (Reiman 1958) (see e.g. [here][1]), which states that the number of edges in an $N$-node graph is at most
$$
\frac{N}{4} ( 1 + \sqrt{4N-3} ).
$$
The Tanner graph on $n$ bits and $k$ parity checks has $N = n+k$ nodes. If it has uniform column weights $d$ then it has $dn$ edges, which gives you the claimed bound.

I am not sure whether the bound can be improved for the special case of bipartite graphs, as is the case for Tanner graphs.

*(Reiman 1958) Reiman, Istvan. "Über ein problem von K. Zarankiewicz." Acta mathematica hungarica 9.3-4 (1958): 269-273.*


  [1]: http://math.mit.edu/~fox/MAT307-lecture09-10.pdf