WLOG $x,y\ge0$, or else we can just switch quadrants as needed.
Suppose for the sake of contradiction that for all positive reals $N,M$, there exists a lattice point $(x,y)$ with $x\ge y\ge N$ such that for some $k>M\ln x$, we have $\gcd(x+i,y+j)>1$ for all $1\le i,j\le k$.
Define $p_{ij}$ to be the smallest prime divisor of $\gcd(x+i,y+j)$ for $1\le i,j\le k$, and let $P$ be the least common multiple of $\{p_{ij}\}_{1\le i,j\le k}$. If $p\le k$ is a prime, then $p$ divides at most $\left(\frac{k}{p}+1\right)^2$ of the $\gcd(x+i,y+j)$, and if $p>k$ then $p$ divides at most $1$ of them. Hence if $X$ is the number of primes larger than $k$ dividing $P$, we have by assumption that
\[X+\pi(k)+2k\ln\ln k+\frac{k^2}{2}\ge X+\sum_{p\le k}\left(1+\frac{k}{p}\right)^2\ge k^2,\]i.e. $X\ge Ck^2$ for some $C>0$ (if we take $M,N$ sufficiently large). But each prime larger than $k$ dividing $P$ is counted exactly once in $X$ and divides $x+i$ for some $1\le i\le k$, so
\[(x+k)^k>(x+1)\cdots(x+k)>k^X\ge k^{Ck^2}\]and therefore
\[e^k+k>x^M+k>x+k>k^{Ck},\]a contradiction for sufficiently large $M$ (note that $k>M$).