Since the explanation was a little more complicated than I initially thought, I figured it would be worth it to combine my comments (and info from Physics SE) into an answer.
Quantum particles satisfy Fermi–Dirac or Bose–Einstein statistics depending on whether they are fermions or bosons. These distributions have the form $$\langle n_i\rangle=\frac{1}{\exp[(\epsilon_i-\mu)/k_bT]\pm1}$$ where the plus/minus is for fermions/bosons.
To consider the high temperature limit, we need to note that not only is there a direct temperature dependence, but $\mu$ is also dependent on temperature. Specifically, in the high temperature limit, $\mu<0$ and $|\mu|>k_bT$. Combining this information, we can make the (very accurate) approximation $$\exp[(\epsilon_i-\mu)/k_bT]\pm1\approx\exp[(\epsilon_i-\mu)/k_bT]$$ as the exponential function will be much larger than $1$. This gives that at high temperatures $$\langle n_i\rangle\approx\frac{1}{\exp[(\epsilon_i-\mu)/k_bT]}$$ which matches the form of the Maxwell–Boltzmann distribution. We can see that at low temperatures these two distributions would not agree, as the additional term in the denominator would become more significant as the exponential got smaller.
It's important to remember that the particles are always indistinguishable; all we have done in the high temperature limit is made an approximation that simplified the functional form. We should not take this coincidental agreement with the MB distribution (for which the particles are assumed to be distinguishable) to imply that the particles have become distinguishable.
