The equations are equivalent to $x^2+t^3=y^2+z^3$ and $x^2+t^3=y^2+z^3+1$. Let $f(n)$ denote the number of solutions to $a^2+b^3=n$ with $0\le a,b\le 10^6$, and set $k=10^{12}+10^{18}$. Then \[N-M=\sum_{n=0}^{k}f(n)^2-\sum_{n=1}^{k}f(n)f(n-1)=\frac{1}{2}\sum_{n=1}^{k}[f(n)-f(n-1)]^2>0,\]since $1=f(0)\ne f(1)=2$.

This one scared me!

http://www.artofproblemsolving.com/community/c3046h1046697_3