 Square-count language is not context-free

The language $\{a^{m^2}{b}^{m^2}:m\ge 0\}$ is not context-free. Pump a word with both block lengths $p^2$. Unequal changes to the blocks destroy equality; an equal positive change $d\le p$ makes both lengths $p^2+d$, which lies strictly between consecutive squares $p^2$ and $(p+1{)}^2$.