To complement Mohan's answer, it is worth noting that there are counterexamples when $R$ contains a field $k$ of prime characteristic $p$.  Indeed, when $p\geq 3$, let $R=k[\![X,Y]\!]/(X^p, Y^p)$, and denote the images of $X$, $Y$ in $R$ by $x$, $y$ respectively.  Then I claim that $t=p$ but $e\geq 2p-2>p$.  To see this, note that any element of $f\in\mathfrak m$ is of the form $f=xg+yh$, and then by Freshman's Dream, $f^p = x^p g^p + y^p h^p = 0$, whereas clearly $x^{p-1} \neq 0$, showing that $t=p$.  On the other hand, $0 \neq x^{p-1} y^{p-1} \in {\mathfrak m}^{2p-2}$.

A characteristic 2 counterexample is given by $k[\![X,Y]\!]/(X^4, Y^4)$ ($k$ any field of char 2), in which case $t=4$ but $e\geq 6$.

To summarize, your question has a 'yes' answer if you are willing to assume the ring contains $\mathbb Q$, but can be 'no' if $R$ contains a field of any other characteristic.  I don't know what happens in mixed characteristic.