The probabilities of getting a straight flush, 4 of a kind, full house, ... , and so on, when dealt 5 cards from a randomly shuffled deck are easily calculated as straight flush .000015, four of a kind .00024, full house .00144, ... and so on.
However, in holdem you have 7 cards with which to make your holdem hand.
Then I should expect to see the above probabilities scaled up by a factor of 21, since I now have 7 choose 5 ways in which to make my hand?
Is this correct?
No, that’s wrong. The most straightforward way to see that it’s wrong is that the probabilities still have to add up to $1$. Perhaps you meant that the probability for nothing (i.e. high card) isn’t scaled up but rather decreases to make up for the increase in everything else; but that also can’t work because the probability for one pair is already more than $\frac1{21}$.
The reason this doesn’t work is that you can have various combinations in those $21$ five-card hands and only the highest one counts, but your approach assumes that all the ones that aren’t the highest (or duplicate the highest) also count.