$ 49 + 20\sqrt [3]{6} = (\sqrt [3]{48} + \sqrt [3]{288} - 1)^2$

You also have to prove there are no others... and a sketch how you found them would be nice.

I try to find $ m,n\in Z^ +$ so that $ 49 + 20\sqrt [3]{6} = (m\sqrt [3]{6} + n\sqrt [3]{36} - 1)^2$ and I have already found them! Now $ \sqrt [3]{48} + \sqrt [3]{288} - 1 = \sqrt [3]{a} + \sqrt [3]{b} - 1\Rightarrow a = 48,b = 288$ or $ a = 288,b = 48$ (Because $ \sqrt [3]{48} + \sqrt [3]{288} - 1\neq 1 - \sqrt [3]{a} - \sqrt [3]{b}$)

In fact the original question (I think which this question maybe should have been) is to find a pair $(a,b)$, rather than find all pairs $(a,b)$. The trick is to set $a=6x^3$ and $b=36y^3$. After expanding we have $x^2\sqrt[3]{36}+6y^2\sqrt[3]{6}+12xy=2y\sqrt[3]{36}+(2x+20)\sqrt[3]{6}+48$. So to find a solution, we look to equate the coefficients: $x^2=2y,6y^2=2x+20,12xy=48$. The pair $(2,2)$ can be shown to satisfy these, hence we obtain tdl's solution $(a,b)=(48,288)$ and it's permutation. If any other solution $(c,d)\not= (a,b)$ exists then it satisfies $\sqrt[3]{c}+\sqrt[3]{d}=\sqrt[3]{a}+\sqrt[3]{b}=2\sqrt[3]{6}+2\sqrt[3]{36}$. This is equivalent to finding all $1\le c\le 277$ for which $6\sqrt[3]{36c^2}+6\sqrt[3]{6c^2}-12\sqrt[3]{36c}-72\sqrt[3]{6c}-144\sqrt[3]{c}+144\sqrt[3]{6}+144\sqrt[3]{6}$ is an integer, I have an inkling it is only $c=48$. Perhaps there exists a nicer method.