Commuting positive operators with equal squares are equal ($ST=TS,S^2=T^2⟹S=T$)

For bounded positive operators satisfying the hypotheses, $(S-T)(S+T)=0$, so $S-T$ vanishes on the closure of the range of $S+T$. The kernel of a sum of positive operators identity shows that it also vanishes on $\ker (S+T)$. Self-adjointness and the Hilbert projection theorem decompose the whole Hilbert space as that closed range plus that kernel. Thus $S-T=0$. This proves the commuting case of uniqueness of a positive square root without using a spectral representation.