How does that work? You have two 100gb drives in raid1, 80% full, you replace one with a 200gb disk and write 50gb to the array - how is your 130gb of data protected against either drive failing?
For two devices, 1x redundancy (so 2x copies of everything) will always limit your storage to the size of the smaller device otherwise it is not possible to have two copies of everything you need to store. As soon as you add a third device of at leats 100gb (or replace the 100gb device with one at least 200gb) the other 100gb of your second device will immediately come into play.
Uneven device size support is most useful when:
♯ You have three or more devices, or plan to grow onto three or more from an initial pair.
♯ You want flexibility rwt array growth (support for uneven devices usually (but not always) comes with better support for dynamic array reshaping).
♯ You want better quick repair flexibility: if 4Tb drive fails, you can replace it with 2x2Tb if you don't have a working 4Tb unit on-hand.
♯ You want variable redundancy (support for uneven devices sometimes comes with support for variable redundancy: keeping 3+ copies of important data, or data you want to access fastest via striping of reads, 2 copies of other permanent data, and 1 copy of temporary storage, all in the same array). In this instance the “wasted” part of the 200gb drive in your example could be used for scratch data designated as not needing to be stored with redundancy.
e.g. you have 3 100GB drives, total capacity in raid 1 is 150GB.
If you replace a broken one with a 200GB one, the total capacity will be increased to 200GB.
You can then extend the pool by adding more such partition-based vdevs as you replace disks, just add the new partitions and add new vdevs.
So if you have a 1TB disk, a 2TB disk and a 4 TB disk, you could have mirrors (d1p1,d3p1), (d2p1,d3p2), (d2p2,d3p3) with a total of 3TB mirrored, and 1 TB available. If you swap the 1TB for a 2TB disk, partition it, replacing the old d1p1 partition with the new and resilver, and once that's done you can add the mirror (d1p2,d3p4) and get the full 4TB redundant storage.
Not a great solution though, as it requires a lot of manual work, and especially write performance will suffer because ZFS will treat the vdevs as being separate and issue IOs in parallel to them, overloading the underlying devices.