Yes, take for example $ a =\sqrt{3}$ and $ b =\sqrt{6}$ and assume that $ \lfloor a^{m}\rfloor =\lfloor b^{n}\rfloor = k$ for some positive integers $ m,n,k$. Then we need $ k^{2}\le 3^{m},6^{n}< (k+1)^{2}$, so that $ |3^{m}-6^{n}|\le 2k$. Since obviously $ m > n$ we have that $ 2k\ge 3^{n}.|3^{m-n}-2^{n}|\ge 3^{n}$. But then $ 4.6^{n}\le 4k^{2}\le 9^{n}$, so $ 4\le\left(\frac{3}{2}\right)^{n}$, so $ n\le 3$. But $ (\lfloor b^{n}\rfloor) = 2,6,14,...$, while $ (\lfloor a^{m}\rfloor) = 1,3,5,9,15,...$, so no terms are equal, QED.

Peter wrote: Yes, take for example $ a = \sqrt {3}$ and $ b = \sqrt {6}$ and assume that $ \lfloor a^{m}\rfloor = \lfloor b^{n}\rfloor = k$ for some positive integers $ m,n,k$. Then we need $ k^{2}\le 3^{m},6^{n} < (k + 1)^{2}$, so that $ |3^{m} - 6^{n}|\le 2k$. Since obviously $ m > n$ we have that $ 2k\ge 3^{n}.|3^{m - n} - 2^{n}|\ge 3^{n}$. But then $ 4.6^{n}\le 4k^{2}\le 9^{n}$, so $ 4\le\left(\frac {3}{2}\right)^{n}$, so $ n\le 3$. But $ (\lfloor b^{n}\rfloor) = 2,6,14,...$, while $ (\lfloor a^{m}\rfloor) = 1,3,5,9,15,...$, so no terms are equal, QED. Sorry I don't quite catch for the statement $ k^{2}\le 3^{m},6^{n} < (k + 1)^{2}$, so that $ |3^{m} - 6^{n}|\le 2k$ But we can choose $ a = 2 + \sqrt {3}$ $ b = \frac {5 + \sqrt17}{2}$ easily prove $ \lfloor a^{m}\rfloor$is odd and $ \lfloor b^{n}\rfloor$is even.Leads to contradiction.