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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 of equality 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.

EDIT: Equality fails in any mixed characteristic $(p^c, p)$. To see this, let $A := {\mathbb Z}/(p^c)$ and $R := A[X,Y]/(X^p, Y^p)$. First note that $0\neq p^{c-1} (xy)^{p-1} \in {\mathfrak m}^{c+2p-3}$, whence $e>c+2p-3$. However, I claim that $t \leq c+2p-3$. To see this, note that any element of $\mathfrak m$ has the form $pf+xg+yh$. We have $(xg+yh)^{2p-1}=0$ since every term in the expansion is divisible by $x^p$ or $y^p$, and by a similar computation we have $$ (xg+yh)^{2p-2} = {2p-2 \choose p-1} (xygh)^{p-1}. $$ We have $$ (pf+xg+yh)^{c+2p-3} = \sum_{i=0}^{c+2p-3} {c+2p-3 \choose i} (pf)^i (xg+yh)^{c+2p-3-i}, $$ and by the above considerations, the only term that potentially survives is the term where $i=c-1$. That is, $$ (pf+xg+yh)^{c+2p-3} = {c+2p-3 \choose c-1} (pf)^{c-1} (xg+yh)^{2p-2} = {c+2p-3 \choose c-1} (pf)^{c-1} {2p-2 \choose p-1} (xygh)^{p-1}. $$ But it is elementary to check that $p \divides {2p-2 \choose p-1}$, whence $p^c$ divides the displayed term, which is then $0$ in $R$.

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 of equality 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.

EDIT: Equality fails in any mixed characteristic $(p^c, p)$. To see this, let $A := {\mathbb Z}/(p^c)$ and $R := A[X,Y]/(X^p, Y^p)$. First note that $0\neq p^{c-1} (xy)^{p-1} \in {\mathfrak m}^{c+2p-3}$, whence $e>c+2p-3$. However, I claim that $t \leq c+2p-3$. To see this, note that any element of $\mathfrak m$ has the form $pf+xg+yh$. We have $(xg+yh)^{2p-1}=0$ since every term in the expansion is divisible by $x^p$ or $y^p$, and by a similar computation we have $$ (xg+yh)^{2p-2} = {2p-2 \choose p-1} (xygh)^{p-1}. $$ We have $$ (pf+xg+yh)^{c+2p-3} = \sum_{i=0}^{c+2p-3} {c+2p-3 \choose i} (pf)^i (xg+yh)^{c+2p-3-i}, $$ and by the above considerations, the only term that potentially survives is the term where $i=c-1$. That is, $$ (pf+xg+yh)^{c+2p-3} = {c+2p-3 \choose c-1} (pf)^{c-1} (xg+yh)^{2p-2} = {c+2p-3 \choose c-1} (pf)^{c-1} {2p-2 \choose p-1} (xygh)^{p-1}. $$ But it is elementary to check that $p \mid {2p-2 \choose p-1}$, whence $p^c$ divides the displayed term, which is then $0$ in $R$.

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.

EDIT: The statement fails in any mixed characteristic $(p^c, p)$. To see this, let $A := {\mathbb Z}/(p^c)$ and $R := A[X,Y]/(X^p, Y^p)$. First note that $0\neq p^{c-1} (xy)^{p-1} \in {\mathfrak m}^{c+2p-3}$, whence $e>c+2p-3$. However, I claim that $t \leq c+2p-3$. To see this, note that any element of $\mathfrak m$ has the form $pf+xg+yh$. We have $(xg+yh)^{2p-1}=0$ since every term in the expansion is divisible by $x^p$ or $y^p$, and by a similar computation we have $$ (xg+yh)^{2p-2} = {2p-2 \choose p-1} (xygh)^{p-1}. $$ We have $$ (pf+xg+yh)^{c+2p-3} = \sum_{i=0}^{c+2p-3} {c+2p-3 \choose i} (pf)^i (xg+yh)^{c+2p-3-i}, $$ and by the above considerations, the only term that potentially survives is the term where $i=c-1$. That is, $$ (pf+xg+yh)^{c+2p-3} = {c+2p-3 \choose c-1} (pf)^{c-1} (xg+yh)^{2p-2} = {c+2p-3 \choose c-1} (pf)^{c-1} {2p-2 \choose p-1} (xygh)^{p-1}. $$ But it is elementary to check that $p \divides {2p-2 \choose p-1}$, whence $p^c$ divides the displayed term, which is then $0$ in $R$.

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 of equality 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.

EDIT: Equality fails in any mixed characteristic $(p^c, p)$. To see this, let $A := {\mathbb Z}/(p^c)$ and $R := A[X,Y]/(X^p, Y^p)$. First note that $0\neq p^{c-1} (xy)^{p-1} \in {\mathfrak m}^{c+2p-3}$, whence $e>c+2p-3$. However, I claim that $t \leq c+2p-3$. To see this, note that any element of $\mathfrak m$ has the form $pf+xg+yh$. We have $(xg+yh)^{2p-1}=0$ since every term in the expansion is divisible by $x^p$ or $y^p$, and by a similar computation we have $$ (xg+yh)^{2p-2} = {2p-2 \choose p-1} (xygh)^{p-1}. $$ We have $$ (pf+xg+yh)^{c+2p-3} = \sum_{i=0}^{c+2p-3} {c+2p-3 \choose i} (pf)^i (xg+yh)^{c+2p-3-i}, $$ and by the above considerations, the only term that potentially survives is the term where $i=c-1$. That is, $$ (pf+xg+yh)^{c+2p-3} = {c+2p-3 \choose c-1} (pf)^{c-1} (xg+yh)^{2p-2} = {c+2p-3 \choose c-1} (pf)^{c-1} {2p-2 \choose p-1} (xygh)^{p-1}. $$ But it is elementary to check that $p \divides {2p-2 \choose p-1}$, whence $p^c$ divides the displayed term, which is then $0$ in $R$.

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.

EDIT: Okay, now I know at least one counterexample in mixed characteristic as well. Let $A := {\mathbb Z}/(4)$, and let $R := A[\![X,Y]\!]/(X^2, Y^2)$. Then $t=3$ but $e>3$. To see that $e>3$, consider the nonzero element $2xy \in {\mathfrak m}^3$. To see that $t=3$, first note that $t>2$ because $(x+y)^2 = 2xy \neq 0$. On the other hand, any element of $\mathfrak m$ has the form $2f +xg+yh$. We have \begin{align*} (2f+(xg+yh))^3 &= (2f)^3 + 3(2f)^2 (xg+yh)+3(2f)(xg+yh)^2 + (xg+yh)^3\\ &= 4 \cdot 2f^3+4 \cdot 3f^2(xy+gh) + 4 \cdot 3fxgyh + (xg+yh)^3\\ &= (xy+yh)^3, \end{align*} so it suffices to see that any element of the form $(xg+yh)^3$ vanishes. But by the pigeonhole principle, each term of the expansion has either an $x^2$ or a $y^2$, and is hence 0.

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.

EDIT: The statement fails in any mixed characteristic $(p^c, p)$. To see this, let $A := {\mathbb Z}/(p^c)$ and $R := A[X,Y]/(X^p, Y^p)$. First note that $0\neq p^{c-1} (xy)^{p-1} \in {\mathfrak m}^{c+2p-3}$, whence $e>c+2p-3$. However, I claim that $t \leq c+2p-3$. To see this, note that any element of $\mathfrak m$ has the form $pf+xg+yh$. We have $(xg+yh)^{2p-1}=0$ since every term in the expansion is divisible by $x^p$ or $y^p$, and by a similar computation we have $$ (xg+yh)^{2p-2} = {2p-2 \choose p-1} (xygh)^{p-1}. $$ We have $$ (pf+xg+yh)^{c+2p-3} = \sum_{i=0}^{c+2p-3} {c+2p-3 \choose i} (pf)^i (xg+yh)^{c+2p-3-i}, $$ and by the above considerations, the only term that potentially survives is the term where $i=c-1$. That is, $$ (pf+xg+yh)^{c+2p-3} = {c+2p-3 \choose c-1} (pf)^{c-1} (xg+yh)^{2p-2} = {c+2p-3 \choose c-1} (pf)^{c-1} {2p-2 \choose p-1} (xygh)^{p-1}. $$ But it is elementary to check that $p \divides {2p-2 \choose p-1}$, whence $p^c$ divides the displayed term, which is then $0$ in $R$.