The complete graph $K_{14}$ is much bigger than needed for this result; here are three proper subgraphs with the same property.
I. Every $2$-coloring of the complete bipartite graph $K_{3,7}$ contains a monochromatic $C_4$.
Let $K_{3,7}$ have partite sets $V_1,V_2$ with $|V_1|=3$ and $|V_2|=7$. Each vertex in $V_2$ is joined by edges of one color to two vertices in $V_1$. By the pigeonhole principle, two vertices in $V_2$ are joined by edges of the same color to the same two vertices in $V_1$.
II. Every $2$-coloring of the complete graph $K_6$ contains a monochromatic $C_4$. (This was shown in Misha Lavrov's answer; the following argument is perhaps slightly simpler.)
We may assume that there is a vertex $x$ which is incident with at most two blue edges. In fact, we may assume that $x$ is incident with exactly two blue edges; otherwise $x$ would be incident with four red edges, call them $xa_1$, $xa_2$, $xa_3$, $xa_4$, and then we could consider a vertex $y\notin\{x,a_1,a_2,a_3,a_4\}$ and proceed as in paw88789's answer. (It may be even easier to observe that the subgraph induced by $\{a_1,a_2,a_3,a_4\}$ contains either a red $P_3$ or a blue $C_4$.)
So let $x$ be incident with exactly two blue edges, $xy$ and $xz$. If one of the remaining three vertices is joined by blue to both $y$ and $z$, then we have a blue $C_4$. On the other hand, if each of those three vertices is joined by red to a vertex in $\{y,z\}$, then two of them are joined by red to the same vertex in $\{y,z\}$, and also to $x$, making a red $C_4$.
III. Every $2$-coloring of the complete bipartite graph $K_{5,5}$ contains a monochromatic $C_4$.
The proof is left as an exercise for the reader.
Here's a generalization of the $(k+1) \times (\frac{k^2(k+1)}{2}+1)$ proof to square grids.
Suppose we have an $n \times n$ grid. If a column of the grid has $n_i$ squares of the $i^{\text{th}}$ color, with $n_1 + \dots + n_k = n$, then it has $\binom{n_1}{2} + \dots + \binom{n_k}{2}$ monochromatic pairs of squares, which can never repeat in the same color in a different column. By convexity, this number is at least $k\binom{n/k}{2}$.
The total supply of such pairs is $k \binom n2$: there are $\binom n2$ pairs of squares, and each pair can be monochromatic in $k$ different colors. So the constraint on $n$ is that $nk \binom{n/k}{2} \le k \binom n2$: we can pick $n$ columns without exhausting the supply. This inequality just barely fails for $n = k^2+k$: we get $\frac12 k^2(k+1)(k^2+k)$ on the left and $\frac12 k^2(k+1)(k^2+k-1)$ on the right.
Therefore a $(k^2+k) \times (k^2+k)$ lattice can never be colored, but there is still hope for a $(k^2+k-1) \times (k^2+k-1)$ lattice.
We can destroy this hope and reduce the bound by $1$, incidentally proving that when $k=2$, $4\times 4$ is best possible.
In this $(k^2+k-1) \times (k^2+k-1)$ lattice, the most even way to color a column is to use one color $k$ times and the others $k+1$ times, which uses up $\binom k2 + (k-1)\binom{k+1}{2} = \frac12 k(k^2+k-2)$ monochromatic pairs. We have a total supply of $k \binom{k^2+k-1}{2}$ monochromatic pairs; dividing by $k^2+k-1$, that gives us $\frac 12k(k^2+k-2)$ per column. So the only way we have any chance is to color each column in the most even way possible.
In every column, one color is underused. With $k^2+k-1$ columns, not every color can be underused $k+1$ times; one color is only underused $k$ times. Across all $k^2+k-1$ columns, that color appears in $k\binom k2 + (k^2-1)\binom{k+1}{2} = $ monochromatic pairs, which is more than the $\binom{k^2+k-1}{2}$ monochromatic pairs of that color we are allowed. Some monochromatic pair repeats in that column, giving us a rectangle.
Now we have an upper bound of $(k^2+k-2) \times (k^2+k-2)$, which is possible for $k=2$.
I don't know if this upper bound is tight, but it is the right order of magnitude: at least for all prime powers $k$, we can $k$-color the $k^2 \times (k^2+k)$ lattice. Here is an example when $k=3$:
The idea is to find $k+1$ partitions of ${1,\dots,k^2}$ so that no two elements are in the same partition together. To use a partition to color a column, assign color to each part of the partition, and use that color on every element of that part. Each partition can be used on $k$ columns if we permute the colors assigned to the parts. in the grid above, the first $3$ columns use the partition $\{1,2,3\} \cup \{4,5,6\} \cup \{7,8,9\}$. The next $3$ use the partition $\{1,5,9\} \cup \{2,6,7\} \cup \{3,4,8\}$.
How do we find these $k+1$ partitions? We identify $\{1,\dots,k^2\}$ with $\mathbb F_k^2$, where $\mathbb F_k$ is the field with $k$ elements. The plane $\mathbb F_k^2$ can be partitioned into the set of all lines with a given slope. The slope of a line can be an element of $\mathbb F_k$, or $\infty$ for vertical lines: $k+1$ possibilities. Any two points are in only one line, so they are together in only one partition.
In particular, the $k^2 \times k^2$ square lattice can also be $k$-colored when $k$ is a prime power. So the optimal value is somewhere between $k^2$ and $k^2+k-2$.
Best Answer
Such a rectangle will occur in a grid of $19$ columns by $4$ rows.
By the pigeonhole principle, each column must have a repeated colour point. Ignoring any later repeats, classify the columns according to the first two repeat colour positions; there are $6$ options: $(1,2), (1,3), (1,4), (2, 3), (2, 4)$ and $(3,4)$. So since there are also $3$ colour options for the repeated colours there are only $6\times3=18$ different options for repeated colour by position. Therefore, by the pigeonhole principle again, in a $19\times 4$ grid we must have a suitable rectangle with identically coloured corners.
From an observation by Pere in comments:
For $n$ colours, using the same approach, we would adopt a grid of $(n+1)$ rows and then require $n{n+1 \choose 2}+1 = n(n+1)n/2 +1 = (n^3+n^2)/2 + 1$ columns to find a rectangle with same-coloured corners.