Is it possible to prove EQTM is undecidable by the Rice theorem?
You can reduce it to a problem about single Turing machines by fixing one of the inputs. For example, you can take $M_2$ to be some Turing machine which accepts nothing. This shows that $EQ_{TM}$ is at least as hard as deciding emptiness.

Another approach is to encapsulate the pair of Turing machines $M_1,M_2$ in one Turing machine $M$, which upon input $0x$ will simulate $M_1$ on $x$, and upon input $1x$ will simulate $M_2$ on $x$ (on the empty input it can arbitrarily halt). There is a simple reduction showing that $EQ_{TM}$ is equivalent to the corresponding problem for $M$, and then you can apply Rice's theorem.