Are ACF plots sufficient to detect white noise assuming probable existence of non linear associations?
In other words, are ACF plots sufficient for determining whether a signal is a white noise?

A plot is never sufficient. An impulse-shaped ACF is. (A plot can only ever be an estimate based on observation; to find the true ACF, you'd have to observe the process for infinity, which you can't.) You can only observe that the plot approximates the ideal ACF sufficiently much (to your definition of &quot;sufficient&quot;).
Anyway, assuming you mean &quot;ACF being a dirac delta&quot;, yeah, that fulfills the definition of white. That is sufficient, because it literally is the definition.

can there be a scenario where an ACF plot that is considered to reflect white noise be actually representative of non-linear associations

No, because white noise is defined to be any random process with a delta dirac ACF.

certain non-linear associations can exhibit zero autocorrelations

Yep! &quot;associations&quot; is a bit of very vague term, though. I'd prefer if you said:

Are there discrete random processes (i.e. random sequences) where elements aren't independent from each other, but still be uncorrelated?

Independence is a strictly defined term: It means that the joint density functions of multiple points in time is the product of the individual densities at these times. I.e.: Let $X(t, \xi)\in \mathbb C, \, t \in \mathbb Z$ be a process at times $t$ (with different realisations $\xi$).
Then, the ACF is defined as
$$\phi_{XX}(t_1,t_2) = \mathbb E\left(
X(t_1)X^*(t_2)
\right)$$
and a process is white iff (assuming wide-sense stationarity, so that $\phi(t, t+\tau) =: \phi(\tau) \forall t$)
$$\phi_{XX}(t_1,t_2)=\phi_{XX}(\tau)=\begin{cases}
&gt; 0 &amp; t_1=t_2\iff \tau=0\\
= 0 &amp; t_1\ne t_2\iff \tau \ne 0
\end{cases},$$
from which it's easy to see that different elements from an observed series are uncorrelated.
However, as you probably know, correlation is not the same as independence.
Counterexample:
$$Y(t) = \begin{cases}
U(t) &amp; t\ge 0\\
S\cdot Y(-t) &amp; t &lt; 0
\end{cases},$$
with $U(t)\sim\mathcal U([-1,1])$ (independently uniformly drawn from the $[-1,1]$ interval), and $S\sim\mathcal U(\{-1,1\})$ (a randomly independently drawn sign).
Obviously, different elements from the sequence are not independent. $Y(-5)$ is either $Y(5)$ or $-Y(5)$. Other values are not allowed.
Checking the ACF, you'll see
\begin{align}
\phi_{XX}(t_1,t_2) &amp;= \mathbb E\left(
X(t_1)X^*(t_2)
\right)\\
&amp;=\begin{cases}
\mathbb E\left(
U^2
\right) =\frac13&amp;\text{for }t_1=t_2&amp;\text{(Variance of uniform continous RV)}\\
0 &amp;\text{for } t_1\ne -t_2 &amp;\text{(the $U(t)$ are uncorr.)}\\
\mathbb E\left(
S\cdot U^2
\right)
 &amp;\text{for } -t_1=t_2
\end{cases}\\
&amp;=\begin{cases}
\frac13&amp;\text{for }t_1=t_2\\
0 &amp;\text{for } t_1\ne -t_2\\
\mathbb E(S)\mathbb E\left(
U^2
\right)
 &amp;\text{for } -t_1=t_2 &amp; \text{(Since $S(t)$,$U(t)$ indep.)}
\end{cases}\\
&amp;=\begin{cases}
\frac13&amp;\text{for }t_1=t_2\\
0 &amp;\text{for } t_1\ne -t_2\\
0\cdot\frac13
 &amp;\text{for } -t_1=t_2 \\
\end{cases}\\
&amp;=\begin{cases}
\frac13&amp;\text{for }t_1=t_2\\
0 &amp;\text{for } t_1\ne t_2
\end{cases},
\end{align}
which seems to say &quot;this process is white&quot;.
BUT: this process is not Wide-Sense Stationary, so applying the term &quot;white&quot; to it doesn't &quot;work&quot;.  So, you need to be very careful to what you apply the label &quot;white&quot;, to begin with.