"The fourth moment is equal to three times the squared second moment."
$$\langle X^4 \rangle = 3\langle X^2 \rangle^2$$
Is this generally true for any distribution?
I came across this equality while studying Binder cumulants in statistical physics and couldn't find any further explanations in addition to the sentence above.
First: kurtosis is not the fourth moment, but the fourth moment divided by the squared second moment (kind of normalization):
$$ \text{Kurt}(X)= \frac{E[(X-\mu)^4]}{(E[(X-\mu)^2])^2}=\frac{m_4}{m_2^2}=\frac{m_4}{\sigma^4}$$
Second: there is also the excess kurtosis, defined as $\kappa'=\text{Kurt}(X)-3$
Your question amounts to ask if $\kappa'=0$ always. The answer is, of course, no.
As the linked Wikipedia article says, the distributions where $\kappa'=0$ are called mesokurtic, of which the normal distribution is the main (but not only) example.
But the excess kurtosis can also be negative (eg, exponential distribution) or positive (eg: uniform distribution).