As @ttnphns and @nick-cox said, SVD is a numerical method and PCA is an analysis approach (like least squares). You can do PCA using SVD, or you can do PCA doing the eigen-decomposition of $X^T X$ (or $X X^T$), or you can do PCA using many other methods, just like you can solve least squares with a dozen different algorithms like Newton's method or gradient descent or SVD etc.
So there is no "advantage" to SVD over PCA because it's like asking whether Newton's method is better than least squares: the two aren't comparable.
Christoph Hanck has not posted the details of his proposed example. I take it he means the uniform distribution on the interval $[0,\theta],$ based on an i.i.d. sample $X_1,\ldots,X_n$ of size more than $n=1.$
The mean is $\theta/2$.
The MLE of the mean is $\max\{X_1,\ldots,X_n\}/2.$
That is biased since $\Pr(\max < \theta) = 1,$ so $\operatorname{E}({\max}/2)<\theta/2.$
PS: Perhaps we should note that the best unbiased estimator of the mean $\theta/2$ is not the sample mean, but rather is $$\frac{n+1} {2n} \cdot \max\{X_1,\ldots,X_n\}.$$ The sample mean is a lousy estimator of $\theta/2$ because for some samples, the sample mean is less than $\dfrac 1 2 \max\{X_1,\ldots,X_n\},$ and it is clearly impossible for $\theta/2$ to be less than ${\max}/2.$
end of PS
I suspect the Pareto distribution is another such case. Here's the probability measure:
$$
\alpha\left( \frac \kappa x \right)^\alpha\ \frac{dx} x \text{ for } x >\kappa.
$$
The expected value is $\dfrac \alpha {\alpha -1 } \kappa.$ The MLE of the expected value is
$$
\frac n {n - \sum_{i=1}^n \big((\log X_i) - \log(\min)\big)} \cdot \min
$$
where $\min = \min\{X_1,\ldots,X_n\}.$
I haven't worked out the expected value of the MLE for the mean, so I don't know what its bias is.
Best Answer
The MLE is more efficient when the distributional assumptions are correctly specified. For instance, if the linear model satisfies, $Y = b X + \epsilon$ where $\epsilon$ comes from, say, a shifted 0 meanWeibull distribution, the MLE will account for the skewness of the errors and the LS will not. It happens to be the case that when $\epsilon$ is normally distributed, you arrive at the least squares estimator anyway.