We start with some lemmas.

Lemma. For positive integers $m > 3$ and $k > 2$ we have $m^{m+k} > (m+k+1)^m$.
Proof. We must show that $m^k > \left(\frac{m+k+1}{m}\right)^m = \left(1 + \frac{k+1}{m}\right)^m$. But we have $\left(1 + \frac{k+1}{m}\right)^m < e^{k+1}$ and so
$$\left(1 + \frac{k+1}{m}\right)^m < e^{k+1} < (2.8)^{k+1} \le (2.8)^{\frac{4}{3}k} < 4^k \le m^k.$$

Lemma. For all positive integers $m > 4$ and $n > 5$, we have $e^m > (m+1)^3$ and $2^n > (n+1)^2$.
Proof. The proof of both parts are consequences of some elementary inductions.
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Now for the main problem suppose that $k$ is the largest possible integer and $x_1 < x_2 < \dots < x_k$ are $k$ numbers satisfying the condition. First note that if $[x_i] = [x_j]$ for some $i, j$ we have
$$x_i^{[x_i]} = x_j^{[x_j]} \implies x_i = x_j \implies i = j$$
This means that $[x_i]$'s are distinct and so $[x_k] - [x_i] \ge k - i$. We claim that $k \le 4$. Assume to the contrary that $k > 4$.

* If $[x_2] > 3$ since $[x_k] - [x_2] > 2$ using the first lemma we get $([x_k] + 1)^{[x_2]} < [x_2]^{[x_k]}$. But on the other hand we have
$$[x_2]^{[x_k]} \le x_2^{[x_k]} = x_k^{[x_2]} \le ([x_k] + 1)^{[x_2]}$$
This leads to a contradiction and implies that $[x_1] < [x_2] \le 3$.

* If $[x_2] = 3$, then
$$3^{[x_k]} \le x_2^{[x_k]} = x_k^{[x_2]} = x_k^3 \le ([x_k] + 1)^3$$
By the second lemma, we deduce that $[x_k] \le 4$ and therefore $k \le 4$ which contradicts with our assumption.

* If $[x_2] = 2$, then similar to the previous part we have
$$2^{[x_k]} \le x_2^{[x_k]} = x_k^{[x_2]} = x_k^2 \le ([x_k] + 1)^2$$
and so by the second lemma $[x_k] \le 5$ therefore $k \le 5$. Now if $k = 5$, then $[x_i] = i$ for all $1 \le i \le 5$ which contradicts, because
$$6 > x_5 = x_1^5 = x_1^2 \times x_1^3 = x_2 \times x_3 \ge 2 \times 3 = 6$$
Therefore $k \le 4$. For $k = 4$, the numbers $x_1 = 3^{\frac{1}{3}}$, $x_2 = 3^{\frac{2}{3}}$, $x_3 = 3$, $x_4 = 3^{\frac{4}{3}}$ satisfy the condition, because $[x_i] = i$ and so $x_i^{[x_j]} = 3^{\frac{ij}{3}} = x_j^{[x_i]}$. So the proof is complete and the desired maximum is $4$. ■