I was trying to solve a different problem of representation across multiple bases, looking at this page http://primefan.tripod.com/BaseReps.html when I noticed this: $$11000100, 21021, 3010, 1241, 524, 400, 304, 237, 196, 169, 144, 121, 100, \textrm{D1}, \textrm{C4}$$ (I added commas to make it more readable)
Meaning that $196$ in base $11$ reads $169$, and in decimal $169 = 13^2$; $196$ in base $12$ reads $144$, and in decimal $144 = 12^2$; $196$ in base $13$ reads $121$, and in decimal $121 = 11^2$; $196$ in base $14$ reads $100$, and in decimal $100 = 10^2$; then for base $15$ this pattern stops.
Looking at the chart, I notice something similar for $25$ in bases $4$ and $5$; for $36$ ($5$ and $6$); for $49$ ($5$ to $7$); $64$ ($6$ to $8$); $81$ ($7$ to $10$).
The chart is limited in that it only goes from binary to hexadecimal, and I am limited in that I'm only looking at these squares from a decimal perspective, but maybe it gets broader if we recognize squares from the perspective of other bases.
Clearly $b^2$ will be represented as $100$ in base $b$. But how to explain when $b^2$ is represented as $121$ in base $b - 1$? Can that explanation be used to find longer runs than what I've found for $196$?
$b^2$ is represented as $121_{b-1}$ because if we let $c=b-1$ we are asking for $(c+1)^2=c^2+2c+1$. As long as $c \gt 2$ so the middle term does not carry this is written as $121_c$. Similarly it will be $144_{b-2}$ as long as $4 \lt b-2$ so there is no carry. This shows that you can get an arbitrarily long run by making the base high enough. Counting down in base from $b^2$ we will have $100, 121, 144, 169, 18(16),$ etc. and until the ones digit carries they will all look like squares. What, you say, $18(16)$ doesn't look like a square? That is because in base $10$ the $16$ carries and we get $196$.