Here is a brute force proof that $SL_3(R)$ does not embed into $SL_2(R)$ for any commutative ring $R$ with 1, inspired by Kevin's universality remarks and Alex's observation that $S_3$ in my comment does not land in $SL_2(R)$.

The claim is that the symmetric group $S_3$ cannot be embedded in $SL_2(R)$ for any $R$: suppose it can be, take 2 general matrices $M=\begin{pmatrix}a&b\cr c&d\end{pmatrix}$ and $T=\begin{pmatrix}e&f\cr g&h\end{pmatrix}$ and consider the relations $M^3=id=T^2$, $M^2T=TM$ and $det M=det T=1$. These are 4+4+4+1+1=14 polynomial relations in 8 variables $a,...,h$, and a Groebner basis computation shows that the ideal they generate is $\langle h^2-1,g,f,h-e,d-1,c,b,a-1\rangle$; in Mathematica this is

$\gt$ M = {{a, b}, {c, d}}; T = {{e, f}, {g, h}}; id = {{1, 0}, {0, 1}};

$\gt$ GroebnerBasis[{M.M.M - id, T.T - id, M.M.T - T.M, Det[T] - 1, Det[M]-1 }]

{-1+h^2,g,f,-e+h,-1+d,c,b,-1+a}

In other words, the relations imply that $a=d=1$ and $b=c=0$ for any $R$, so $M=1$, contradicting the assumption that $S_3\to SL(2,R)$ is injective.

On the other hand, using $$ M=\begin{pmatrix}0&0&1\cr 1&0&0\cr 0&1&0\end{pmatrix},\qquad T=\begin{pmatrix}0&-1&0\cr -1&0&0\cr 0&0&-1\end{pmatrix}, $$ we can embed $S_3$ into $SL(3,R)$ for any ring $R$.

P.S. Hopefully, there is a better proof that $S_3$ does not embed into $SL(2,R)$!

Edit: The Groebner basis computation works over $Z[1/2]$, so this only works for rings $R$ with 2 invertible. Kevin & John: thank you for pointing this out!

Here is a brute force proof that $SL_3(R)$ does not embed into $SL_2(R)$ for any commutative ring $R$ with 1 where 2 is invertible, inspired by Kevin's universality remarks and Alex's observation that $S_3$ in my comment does not land in $SL_2(R)$.

The claim is that the symmetric group $S_3$ cannot be embedded in $SL_2(R)$ for any $R$: suppose it can be, take 2 general matrices $M=\begin{pmatrix}a&b\cr c&d\end{pmatrix}$ and $T=\begin{pmatrix}e&f\cr g&h\end{pmatrix}$ and consider the relations $M^3=id=T^2$, $M^2T=TM$ and $det M=det T=1$. These are 4+4+4+1+1=14 polynomial relations in 8 variables $a,...,h$, and a Groebner basis computation shows that the ideal they generate is $\langle h^2-1,g,f,h-e,d-1,c,b,a-1\rangle$; in Mathematica this is

$\gt$ M = {{a, b}, {c, d}}; T = {{e, f}, {g, h}}; id = {{1, 0}, {0, 1}};

$\gt$ GroebnerBasis[{M.M.M - id, T.T - id, M.M.T - T.M, Det[T] - 1, Det[M]-1 }]

{-1+h^2,g,f,-e+h,-1+d,c,b,-1+a}

In other words, the relations imply that $a=d=1$ and $b=c=0$ for any $R$, so $M=1$, contradicting the assumption that $S_3\to SL(2,R)$ is injective.

On the other hand, using $$ M=\begin{pmatrix}0&0&1\cr 1&0&0\cr 0&1&0\end{pmatrix},\qquad T=\begin{pmatrix}0&-1&0\cr -1&0&0\cr 0&0&-1\end{pmatrix}, $$ we can embed $S_3$ into $SL(3,R)$ for any ring $R$.

P.S. Hopefully, there is a better proof that $S_3$ does not embed into $SL(2,R)$!

Edit: The Groebner basis computation works over $Z[1/2]$, so this only works for rings $R$ with 2 invertible. Kevin & John: thank you for pointing this out!

Here is a brute force proof that $SL_3(R)$ does not embed into $SL_2(R)$ for any commutative ring $R$ with 1, inspired by Kevin's universality remarks and Alex's observation that $S_3$ in my comment does not land in $SL_2(R)$.

The claim is that the symmetric group $S_3$ cannot be embedded in $SL_2(R)$ for any $R$: suppose it can be, take 2 general matrices $M=\begin{pmatrix}a&b\cr c&d\end{pmatrix}$ and $T=\begin{pmatrix}e&f\cr g&h\end{pmatrix}$ and consider the relations $M^3=id=T^2$, $M^2T=TM$ and $det M=det T=1$. These are 4+4+4+1+1=14 polynomial relations in 8 variables $a,...,h$, and a Groebner basis computation shows that the ideal they generate is $\langle h^2-1,g,f,h-e,d-1,c,b,a-1\rangle$; in Mathematica this is

$\gt$ M = {{a, b}, {c, d}}; T = {{e, f}, {g, h}}; id = {{1, 0}, {0, 1}};

$\gt$ GroebnerBasis[{M.M.M - id, T.T - id, M.M.T - T.M, Det[T] - 1, Det[M]-1 }]

{-1+h^2,g,f,-e+h,-1+d,c,b,-1+a}

In other words, the relations imply that $a=d=1$ and $b=c=0$ for any $R$, so $M=1$, contradicting the assumption that $S_3\to SL(2,R)$ is injective.

On the other hand, using $$ M=\begin{pmatrix}0&0&1\cr 1&0&0\cr 0&1&0\end{pmatrix},\qquad T=\begin{pmatrix}0&-1&0\cr -1&0&0\cr 0&0&-1\end{pmatrix}, $$ we can embed $S_3$ into $SL(3,R)$ for any ring $R$.

P.S. Hopefully, there is a better proof that $S_3$ does not embed into $SL(2,R)$!

Here is a brute force proof that $SL_3(R)$ does not embed into $SL_2(R)$ for any commutative ring $R$ with 1, inspired by Kevin's universality remarks and Alex's observation that $S_3$ in my comment does not land in $SL_2(R)$.

The claim is that the symmetric group $S_3$ cannot be embedded in $SL_2(R)$ for any $R$: suppose it can be, take 2 general matrices $M=\begin{pmatrix}a&b\cr c&d\end{pmatrix}$ and $T=\begin{pmatrix}e&f\cr g&h\end{pmatrix}$ and consider the relations $M^3=id=T^2$, $M^2T=TM$ and $det M=det T=1$. These are 4+4+4+1+1=14 polynomial relations in 8 variables $a,...,h$, and a Groebner basis computation shows that the ideal they generate is $\langle h^2-1,g,f,h-e,d-1,c,b,a-1\rangle$; in Mathematica this is

$\gt$ M = {{a, b}, {c, d}}; T = {{e, f}, {g, h}}; id = {{1, 0}, {0, 1}};

$\gt$ GroebnerBasis[{M.M.M - id, T.T - id, M.M.T - T.M, Det[T] - 1, Det[M]-1 }]

{-1+h^2,g,f,-e+h,-1+d,c,b,-1+a}

In other words, the relations imply that $a=d=1$ and $b=c=0$ for any $R$, so $M=1$, contradicting the assumption that $S_3\to SL(2,R)$ is injective.

On the other hand, using $$ M=\begin{pmatrix}0&0&1\cr 1&0&0\cr 0&1&0\end{pmatrix},\qquad T=\begin{pmatrix}0&-1&0\cr -1&0&0\cr 0&0&-1\end{pmatrix}, $$ we can embed $S_3$ into $SL(3,R)$ for any ring $R$.

P.S. Hopefully, there is a better proof that $S_3$ does not embed into $SL(2,R)$!

Edit: The Groebner basis computation works over $Z[1/2]$, so this only works for rings $R$ with 2 invertible. Kevin & John: thank you for pointing this out!