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), 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.

This is tight up to constants for the case where $k = n$ (see e.g. here).

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

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 $O(\sqrt{k})$. This is known as the Kővári–Sós–Turán theorem (see e.g. here), which states that the number of edges in an $(n,k)$-bipartite graph without 4-cycle is at most $$ (n-1)\sqrt{k} + k. $$ If the Tanner graph has uniform column weights $d$, then it has $nd$ edges, which gives the bound on $d$.

This is tight up to constants for the case where $k = n$ (see e.g. here).

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), 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.

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), 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.

This is tight up to constants for the case where $k = n$ (see e.g. here).

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