This is yet another case where you must pay very close attention to the parameterization of a Weibull model.* This doesn't depend on having censored data, as the same issue arises with uncensored data.
The brms package uses a shape, scale parameterization as in the standard R rweibull() function. In the brms vignette on parameterization, the authors use $\alpha$ to represent the shape and $s$ to represent the scale of a Weibull model.
The "Intercept" in a brm() Weibull model, however, is not $\log(s)$, as assumed by the OP in the statement bayes_scale = round(exp(bayes_intercept),3). As the vignette explains, the package authors
use $\mu$ to refer to the main model parameter, which is usually the mean of the response distribution or some closely related quantity. In a regression framework, $\mu$ is not estimated directly but computed as $\mu=g(\eta)$, where $\eta$ is a predictor term ... and $g$ is the response function (i.e., inverse of the link function).
The "Intercept" term returned by brm() is thus $\log (\mu)$, the log of the mean of the distribution. For a given shape $\alpha$ the Weibull scale parameter $s$ is then:
$$ s = \mu/\Gamma(1+1/\alpha),$$
as described in the vignette.
When I followed the code in this question (using set.seed(101) before generating the random Weibull sample), brm() gave an Intercept of 5.238432 and a shape of 1.461953. With the above formula, that gives a scale $s$ of 207.99.
With the same random Weibull sample, survreg() gave an Intercept (here, the log of the scale in the rweibull() parameterization) of 5.335683, for a scale of 207.61. The apparent discrepancy between the maximum-likelihood and Bayesian estimates of scale thus disappears once you recognize that the Intercept in the latter case, from brm(), is $\log(\mu)$, not $\log(s)$.
As I searched the web for similar cases, I found that this definition of the Intercept for a brm() Weibull model isn't always appreciated, as for example on this page. If you generate samples based on a Weibull shape value near 1, as on that page, you wouldn't notice the discrepancy caught by the OP here, as $\Gamma(2)=1$.
*I got the crucial hint for this solution from this page by Riley King, that illustrates frequentist and Bayesian Weibull modeling of both uncensored and censored data. The author used a grid approximation directly on shape and scale for a Bayesian posterior estimate on uncensored data, then moved to brm() for censored data. He read the manual closely enough to recognize that:
The parameters that get estimated by brm() are the Intercept and shape. We can use the shape estimate as-is, but it’s a bit tricky to recover the scale. The key is that brm() uses a log-link function on the mean $\mu$.