T-tests provide info about the margin of mean difference? I have 2 unpaired datasets where the difference of the sample means is quite substantial (with the naked eye). The t-test shows that the difference is significant (p-value=0.007) and I would like to know if I can use the t-test to claim that one dataset provides significantly better results than the other or do I have to say that t-tests prove the difference is significant and use confidence intervals (or some other method) to show that one performs better than the other one?
If we go by the description of the T-test on Wikipedia: "The t-test can be used, for example, to determine if the means of two sets of data are significantly different from each other".
Based on that I believe t-tests can be used only to show if the means are different, we cannot extract info about the margin of the difference or one dataset is better than the other.
 A: 
if I can use the t-test to claim that one dataset provides significantly better results than the other

I wanted to point out that the expression "statistically significant" has been discouraged by the American Statistical Association. (But personally, I don't feel too strongly against it)

use confidence intervals (or some other method) to show that one performs better than the other one

Confidence intervals can be interpreted as the range of values compatible with the observed data given the assumptions. In fact, it's been suggested that they should be renamed compatibility intervals (see Is the interpretation of a "Compatibility Interval" (Greenland, 2019) valid in general? and links therein).
So it seems to me that reporting confidence intervals is appropriate in your case from that little bit I can tell.
I would also note that there is a duality between p-values and confidence intervals since the 95% confidence interval is the range where the p-value becomes greater than 0.05 (or whatever alpha you chose to determine the interval). So the confidence interval is often considered a more informative alternative to the p-value for reporting results.
A: Consider the two independent fictitious normal datasets below.
set.seed(910)
x1 = rnorm(100, 50, 7)
x2 = rnorm(90, 60, 8)

mean(x1)
[1] 49.86995
mean(x2)
[1] 60.04863

The sample means show $\bar X_1 < \bar X_2$ suggesting that $\mu_1$ may be 'significantly' smaller than $\mu_2.$ Welch two-sample t test in R shows that the null hypothesis $H_0: \mu_1 = \mu_2$ is rejected in
favor of $H_a: \mu_1 < \mu_2.$ with P-value near $0.$
More specifically, the one sided CI: shows 95% confidence that
$\mu_1 - \mu_2 \le -8.412.$
t.test(x1, x2, alt="less")

        Welch Two Sample t-test

data:  x1 and x2
t = -9.5241, df = 187.95, p-value < 2.2e-16
alternative hypothesis: 
 true difference in means is less than 0
95 percent confidence interval:
      -Inf -8.412074
sample estimates:
mean of x mean of y 
 49.86995  60.04863 

Nothing here is a 'proof' that $\mu_1 < \mu_2.$ Of course, because
I simulated these data, we know $\mu_1 - \mu_2 = -10.$ However,
in a real statistical application we would not have such exact information.
The best we can do is to
report that sample means show $\bar X_1 =49.87 < \bar X_2 = 60.05$
and to say we are
'reasonably sure' that $\mu_1 - \mu_2 \le -8.412.$
A: If the null hypothesis is $t=0$, then this is testing if the two means are the same or not.  But you can also test $t > a$, then this tests if the margin is larger than some number.
