I agree with both of you. I have a 5090 in my main rig. I also have an Intel ARC B580 in the little portable rig I built last year for ~$1k before prices went nuts. The portable rig can handle any game I have, and I went looking for heavy stuff after getting my hands on the 5090. Already had CyberPunk 2077, but I added Doom the Dark Ages (mandatory RT + optional path tracing), Indiana Jones and the Great Circle (mandatory RT + optional path tracing), and Borderlands 4 (runs like path tracing on the "badass" preset even though it doesn't use RT). All of them run ok on the budget portable rig with a B580. Of course I have to lower settings, use XeSS, etc. but if that's all I had I'd still play them.I don't think that is stubborn, I think that is accurate and that comes from someone who owns one and uses it to game often. Don't get me wrong: I like mine, I'm glad to have one. It is fun and I enjoy getting good frame rates even when turning shit up to ludicrous detail levels in games like Cyberpunk. But is it absolutely NOT necessary. It is a nice luxury. I also game on my laptop which has a 4080 mobile, so like a 4070 or a bit less of performance. That works completely fine, games still look great and I still get to have lots of fun. Yes I prefer the 5090, of course I do, but it isn't necessary to enjoy gaming.
I would liken it to any other luxury like luxury cars: Yes, nice to have if you can (and choose to) afford one but 100% NOT needed.
Comparing a luxury car to a 5090 is an excellent analogy IMHO, at least for gaming. It's nice but totally unnecessary. Of course there are other uses where you might actually need a 5090 or something even more expensive, but for gaming? Nah. Even 8GB cards still work fine even though we shit on them constantly. They're just stuck at lower settings, and it mostly hits textures. They're so common that AAA games will support them for years to come, just at reduced settings.
Personally I won't mind if next gen is delayed a bit. It'll let me have the top card for longer without spending $.