We can proceed similar as you did. Let $n=\lfloor x\rfloor$. From the assumption $\lfloor x\rfloor\{\sqrt x\}=1$ one concludes $\{\sqrt x\}=\frac1n$. Since $\sqrt x$ changes its integral part at integral values, we can conclude that $\lfloor\sqrt n\rfloor=\lfloor\sqrt x\rfloor$, so that
$$\sqrt x=\lfloor\sqrt x\rfloor+\{\sqrt x\}=\lfloor\sqrt n\rfloor+\frac1n\Longrightarrow x=\lfloor\sqrt n\rfloor^2+\frac{2\lfloor\sqrt n\rfloor}{n}+\frac1{n^2}.$$
Now we have to check under which conditions this value for $x$ indeed satisfies the assumption $\lfloor x\rfloor=n$. For this, let $n=k^2+m$ with $k,m\in\mathbb N_0$ and $k$ maximal, i.e. $\lfloor\sqrt n\rfloor=k$. Now if $n\geq5$, then
$$x=k^2+\frac{2k}{n}+\frac1{n^2}=k^2+\frac{2kn+1}{n^2}<k^2+1,$$
so in this case, we must have $m=0$. Then we have $n=k^2<x<k^2+1$, so indeed $\lfloor x\rfloor=n$, hence we found solutions. The cases $n=0,\ldots,4$ are easily checked, and $\frac94$, which you found, is the only solution in these cases.
To summarize, the solutions are $\frac94$ and $x=\lfloor\sqrt n\rfloor^2+\frac{2\lfloor\sqrt n\rfloor}{n}+\frac1{n^2}$ with $n$ a perfect square other than $1$ or $4$, or letting $n=k^2$, $$x=\frac94\quad\text{and}\quad x=k^2+\frac2k+\frac1{k^4},\quad k\in\mathbb N,\ k>2.$$
Solve for $x$ where both: $$\begin{align*}
x\lfloor x\lfloor x\cdot 6\rfloor\rfloor &= 2001 \tag1\\
\lfloor x \rfloor &= 6 \tag2
\end{align*}$$
From the property that $\lfloor a \rfloor \le a$, the LHS of $(1)$ satisfies
$$\begin{align*}
x\lfloor x\lfloor x\cdot 6\rfloor\rfloor &\le x\cdot x\lfloor x\cdot 6\rfloor
\le x\cdot x\cdot x\cdot 6\\
x^3\cdot 6 &\ge 2001\\
x\cdot 6 &\ge \sqrt[3]{6^2\cdot2001} \approx 41.6
\end{align*}$$
From $(2)$,
$$\begin{align*}
6 \le x &< 7\\
x\cdot 6 &< 42
\end{align*}$$
So $\lfloor x\lfloor x\rfloor\rfloor = \lfloor x\cdot 6\rfloor = 41$.
Similarly, from $x\lfloor x\cdot41\rfloor = x\lfloor x\lfloor x\cdot 6\rfloor\rfloor = 2001$,
$$\begin{align*}
x^2 \cdot 41 &\ge x\lfloor x\cdot41\rfloor = 2001\\
x \cdot 41 &\ge \sqrt{41\cdot 2001} \approx 286.4
\end{align*}$$
And from the previous step,
$$\begin{align*}
41.6\ldots \le x \cdot6 &< 42\\
6.9\ldots \le x &< 7\\
284.3\ldots \le x\cdot 41 &< 287
\end{align*}$$
So $\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor = \lfloor x\cdot 41\rfloor = 286$.
One may say this step works, with $\lfloor x\cdot 6\rfloor$ narrowed down to one possibility, only because the lower and upper bounds of $x\cdot 6$ is near. This, I would agree.
This was also observed by some answers and comments in a similar question: find positive real number x that satisfies $2001=x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor$.
For another example by @Sil, to solve for the positive $x$ where $$x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor = 31$$
Here $\lfloor x\rfloor = \left\lfloor\sqrt[4]{31}\right\rfloor = 2$, then on one side $2x\ge \sqrt[3]{2^2\cdot 31} \approx 4.99$ while $2x < 2\cdot 3 = 6$. The possibilities for the next step is $\lfloor x\lfloor x\rfloor\rfloor = \lfloor 2x\rfloor \in\{4,5\}$:
$$\begin{array}{r|c|c}
\lfloor x\lfloor x\rfloor\rfloor = & 4 & 5 \\\hline
\text{Bounds from $x\lfloor x\rfloor$:}\\
x\cdot 2 = x\lfloor x\rfloor \in
& [4.99\ldots, 5)
& [5,6)\\
\overset{/2}{\implies} x \in
& \left[2.49\ldots, \frac52\right)
& \left[\frac52, \frac62\right)\\
\overset{\cdot \lfloor x\lfloor x\rfloor\rfloor}{\implies} x\lfloor x\lfloor x\rfloor\rfloor \in
& [9.97\ldots, 10)
& \left[12.5, 15\right)\\\hline
\text{Lower bound from the given:}\\
31 = x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor \le x^2\lfloor x\lfloor x\rfloor\rfloor =
& 4x^2
& 5x^2\\
x\lfloor x\lfloor x\rfloor\rfloor \ge
& \sqrt{4\cdot31}\approx 11.1
& \sqrt{5\cdot31}\approx 12.4\\\hline
\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor \in
& \emptyset
& \{12,13,14\}
\end{array}$$
The next step even has 3 possibilities:
$$\begin{array}{r|c|c|c}
\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor =
& 12 & 13 & 14\\\hline
\text{Bounds from $x\lfloor x\lfloor x\rfloor\rfloor$:}\\
x\cdot 5 = x\lfloor x\lfloor x\rfloor\rfloor \in
& [12.5, 13)
& [13, 14)
& [14, 15)\\
\overset{/5}{\implies} x \in
& [2.5, 2.6)
& [2.6, 2.8)
& [2.8, 3)\\
\overset{\cdot \lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor}{\implies} x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor \in
& [30, 31.2)
& [33.8, 36.4)
& [39.2, 42)\\\hline
\text{From the given:}\\
x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor =
& 31
& 31
& 31\\
x = \frac{31}{\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor} =
& \frac{31}{12} \approx 2.5\\\hline
\text{Verifications:}\\
\lfloor x\rfloor =
& \lfloor2.5\ldots\rfloor\\
\lfloor x\lfloor x\rfloor\rfloor =
& \lfloor5.1\ldots\rfloor\\
\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor =
& \lfloor12.91\ldots\rfloor
\end{array}$$
So the only positive solution $x$ is $\frac{31}{12}$, which has $\lfloor x\lfloor x\rfloor\rfloor = 5 \ne \left\lfloor\sqrt[3]{2^2\cdot 31}\right\rfloor = 4$.
Best Answer
Extending my comment as a solution, for $x\in(0,1)$, $\lfloor 2^x \rfloor = 1$, $\lfloor 3^x \rfloor = \begin{cases}1, x < \log_3(2)\\ 2, x\ge \log_3(2) \end{cases}$ and $$\lfloor 6^x \rfloor = \begin{cases}1, x < \log_6(2)\\ 2, \log_6(2) \le x< \log_6(3) \\ 3, \log_6(3) \le x< \log_6(4) \\ 4, \log_6(4) \le x< \log_6(5) \\ 5, \log_6(5) \le x< 1 \end{cases}.$$ Therefore, the left side is either $2$ for $x<\log_2(3)$ or $3$ for $x\ge \log_2(3)$. Thus the solution is $$(\log_{6}\left(2\right),\log_{6}\left(3\right)) \cup (\log_{3}\left(2\right),\log_{6}\left(4\right))$$
Edit : From the inequality $6^x \ge 2^x+3^x+1$, we can take floor$()$ both sides as it an increasing function. Now since $\lfloor x+y \rfloor = \lfloor x \rfloor+\lfloor y \rfloor$ or $\lfloor x \rfloor+\lfloor y \rfloor+1$, we get $$\lfloor 6^x \rfloor \ge \lfloor 2^x \rfloor+\lfloor 3^x \rfloor +1 > \lfloor 2^x \rfloor+\lfloor 3^x \rfloor.$$