$\left\{\begin{array}{l}\frac{1}{4^{x}}+\frac{1}{27^{y}}=\frac{5}{6}, \\ \log _{27} y-\log _{4} x \geqslant \frac{1}{6}, \\ 27^{y}-4^{x} \leqslant 1 .\end{array}\right.$
From (2), we know that $x, y$ are both positive real numbers. For convenience, let $27^{y}=a, 4^{x}=b$. From (3), we know $a \leqslant b+1$. Combining with equation (1), we have $\frac{1}{b+1}+\frac{1}{b} \leqslant \frac{1}{a}+\frac{1}{b}=\frac{5}{6}$. Therefore, $5 b(b+1) \geqslant 6(2 b+1)$. Solving this, we get $b \geqslant 2$ or $b \leqslant-\frac{3}{5}$. However, $b=4^{x}>0$, so $b \geqslant 2$, which implies $x \geqslant \frac{1}{2}$. 

Now, from (2), we know $\log _{27} y \geqslant \frac{1}{6}+\log _{4} x \geqslant \frac{1}{6}+\log _{4} \frac{1}{2}=\frac{1}{6}-\frac{1}{2}=-\frac{1}{3}$. Therefore, $y \geqslant 27^{-\frac{1}{3}}=\frac{1}{3}$.

Returning to equation (1), we get $\frac{5}{6}=4^{-x}+27^{-y} \leqslant 4^{-\frac{1}{2}}+27^{-\frac{1}{3}}=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$. Thus, the inequality holds with equality, which requires $x=\frac{1}{2}, y=\frac{1}{3}$. The pair of real numbers that satisfy the conditions is $(x, y)=$ $\left(\frac{1}{2}, \frac{1}{3}\right)$.