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(Edit: I edited a bit. Now the answer should be clearer, simpler and even correct. Thank you Andrei Smolensky for repeating correcting my embarrassing mistakes here.)

I am surprised that this old question was not fully answered yet. The answer is "No" and it is well known in some circles. In fact, a far more general statement holds:

1. Let $R$ and $S$ be rings (commutative with 1). Then any group homomorphism $\text{SL}_3(R)\to\text{SL}_2(S)$ factors via the non-trivial quotient $\text{SK}_1(3,R):=\text{SL}_3(R)/\text{EL}_3(R)$, where $\text{EL}_3(R)$ is the subgroup generated by elementary matrices in $\text{SL}_3(R)$.

Note that $\text{SL}_3(R)$ contains an epimorphic image of $\text{SL}_3(\mathbb{Z})$ (induced by the map $\mathbb{Z}\to R$). It is well known (and easy) that $\mathrm{EL}_3(\mathbb{Z})\simeq \mathrm{SL}_3(\mathbb{Z})$, thus the image of $\text{SL}_3(\mathbb{Z})$ is contained in $\mathrm{EL}_3(R)$, and in fact $\mathrm{EL}_3(R)$ is generated by $\text{SL}_3(\mathbb{Z})$ as a normal subgroup (as you can observe by playing with commutation relation of elementary matrices). Thus (1) is equivalent to:

2. Let $S$ be a ring (commutative with 1). Then any group homomorphism $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(S)$ is trivial.

We now fix a homomorphism as in statement (2) and assume its image is non-trivial. It is standard that there is a maximal ideal $m\lhd S$ and an integer $k$ such that the image of $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(S/m^k)$ is already non-trivial. Note that the the kernel of $\text{SL}_2(S/m^k)\to\text{SL}_2(S/m)$ is nilpotent and that $\text{SL}_3(\mathbb{Z})$ has no non-trivial nilpotent quotients (every elementary matrix is a commutator, thus it is perfect). It follows that $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(S/m)$ is non-trivial. We are left to prove the following statement:

3. Let $k$ be a field. Then any group homomorphism $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(k)$ is trivial.

(here I had before an argument I liked, but I had to replace it by a simpler one.)

Here is a nice exercise:

4. Let $k$ be a field. Then for any group homomorphism $\text{H}(\mathbb{Z})\to\text{SL}_2(k)$, where $\text{H}(\mathbb{Z})$ is the integral Heisenberg group, the image of the center (=commutator group) of $\text{H}(\mathbb{Z})$ consists of scalar matrices.

Hint: Assume the image of a generator of the center is not a scalar matrix and show that $\text{H}(\mathbb{Z})$ is in the Borel, in which every nilpotent group is abelian (you may assume that $k$ is algebraically closed here).

Remark: Actually, the image of the center of $\text{H}(\mathbb{Z})$ will be trivial unless $\text{char}(k)=2$.

To finish up with (3), observe that every elementary matrix in $\text{SL}_3(\mathbb{Z})$ is the center of a conjugate of $\text{H}(\mathbb{Z})$, thus the image of $\text{SL}_3(\mathbb{Z})$ consists of scalar matrices. But $\text{SL}_3(\mathbb{Z})$ is perfect, so this image is trivial.

(Edit: I edited a bit. Now the answer should be clearer, simpler and even correct. Thank you Andrei Smolensky for repeating correcting my embarrassing mistakes here.)

I am surprised that this old question was not fully answered yet. The answer is "No" and it is well known in some circles. In fact, a far more general statement holds:

1. Let $R$ and $S$ be rings (commutative with 1). Then any group homomorphism $\text{SL}_3(R)\to\text{SL}_2(S)$ factors via the non-trivial quotient $\text{SK}_1(3,R):=\text{SL}_3(R)/\text{EL}_3(R)$, where $\text{EL}_3(R)$ is the subgroup generated by elementary matrices in $\text{SL}_3(R)$.

Note that $\text{SL}_3(R)$ contains an epimorphic image of $\text{SL}_3(\mathbb{Z})$ (induced by the map $\mathbb{Z}\to R$). It is well known (and easy) that $\mathrm{EL}_3(\mathbb{Z})\simeq \mathrm{SL}_3(\mathbb{Z})$, thus the image of $\text{SL}_3(\mathbb{Z})$ is contained in $\mathrm{EL}_3(R)$, and in fact $\mathrm{EL}_3(R)$ is generated by $\text{SL}_3(\mathbb{Z})$ as a normal subgroup (as you can observe by playing with commutation relation of elementary matrices). Thus (1) is equivalent to:

2. Let $S$ be a ring (commutative with 1). Then any group homomorphism $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(S)$ is trivial.

We now fix a homomorphism as in statement (2) and assume its image is non-trivial. Let $\mathfrak{n}<S$ denote the nilpotent radical. It is easy to see that every finitely generated subgroup of the kernel of $\text{SL}_2(S)\to \text{SL}_2(S/\mathfrak{n})$ is nilpotent. By the facts that $\text{SL}_3(\mathbb{Z})$ is finitely generated without nilpotent quotients we deduce that it is mapped non-trivially to $\text{SL}_2(S/\mathfrak{n})$. Since $\mathfrak{n}$ is the intersection of all prime ideal we deduce that $\text{SL}_3(\mathbb{Z})$ is mapped non-trivially to $\text{SL}_2(S/\mathfrak{p})$ for some prime ideal $\mathfrak{p}<S$. By letting $k$ be the field of fractions of $S/\mathfrak{p}$ we see that it is enough to prove the following statement:

3. Let $k$ be a field. Then any group homomorphism $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(k)$ is trivial.

(here I had before an argument I liked, but I had to replace it by a simpler one.)

Here is a nice exercise:

4. Let $k$ be a field. Then for any group homomorphism $\text{H}(\mathbb{Z})\to\text{SL}_2(k)$, where $\text{H}(\mathbb{Z})$ is the integral Heisenberg group, the image of the center (=commutator group) of $\text{H}(\mathbb{Z})$ consists of scalar matrices.

Hint: Assume the image of a generator of the center is not a scalar matrix and show that $\text{H}(\mathbb{Z})$ is in the Borel, in which every nilpotent group is abelian (you may assume that $k$ is algebraically closed here).

Remark: Actually, the image of the center of $\text{H}(\mathbb{Z})$ will be trivial unless $\text{char}(k)=2$.

To finish up with (3), observe that every elementary matrix in $\text{SL}_3(\mathbb{Z})$ is the center of a conjugate of $\text{H}(\mathbb{Z})$, thus the image of $\text{SL}_3(\mathbb{Z})$ consists of scalar matrices. But $\text{SL}_3(\mathbb{Z})$ is perfect, so this image is trivial.

(Edit: I edited a bit. Now the answer should be clearer, simpler and even correct. Thank you Andrei Smolensky for repeating correcting my embarrassing mistakes here.)

I am surprised that this old question was not fully answered yet. The answer is "No" and it is well known in some circles. In fact, a far more general statement holds:

1. Let $R$ and $S$ be rings (commutative with 1). Then any group homomorphism $\text{SL}_3(R)\to\text{SL}_2(S)$ factors via the non-trivial quotient $\text{SK}_1(3,R):=\text{SL}_3(R)/\text{EL}_3(R)$, where $\text{EL}_3(R)$ is the subgroup generated by elementary matrices in $\text{SL}_3(R)$.

Note that $\text{SL}_3(R)$ contains an epimorphic image of $\text{SL}_3(\mathbb{Z})$ (induced by the map $\mathbb{Z}\to R$). It is well known (and easy) that $\mathrm{EL}_3(\mathbb{Z})\simeq \mathrm{SL}_3(\mathbb{Z})$, thus the image of $\text{SL}_3(\mathbb{Z})$ is contained in $\mathrm{EL}_3(R)$, and in fact $\mathrm{EL}_3(R)$ is generated by $\text{SL}_3(\mathbb{Z})$ as a normal subgroup (as you can observe by playing with commutation relation of elementary matrices). Thus (1) is equivalent to:

2. Let $S$ be a ring (commutative with 1). Then any group homomorphism $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(S)$ is trivial.

We now fix a homomorphism as in statement (2) and assume its image is non-trivial. It is standard that there is a maximal ideal $m\lhd S$ and an integer $k$ such that the image of $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(S/m^k)$ is already non-trivial. Note that the the kernel of $\text{SL}_2(S/m^k)\to\text{SL}_2(S/m)$ is nilpotent and that $\text{SL}_3(\mathbb{Z})$ has no non-trivial nilpotent quotients (every elementary matrix is a commutator, thus it is perfect). It follows that $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(S/m)$ is non-trivial. We are left to prove the following statement:

3. Let $k$ be a field. Then any group homomorphism $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(k)$ is trivial.

(here I had before an argument I liked, but I had to replace it by a simpler one.)

Here is a nice exercise:

4. Let $k$ be a field. Then for any group homomorphism $H\to\text{SL}_2(k)$, where $H=\text{H}(\mathbb{Z})$ is the integral Heisenberg group, the image of the center (=commutator group) of $H$ consists of scalar matrices.

Hint: Assume the image of a generator of the center is not a scalar matrix and show that $H$ is in the Borel, in which every nilpotent group is abelian (you may assume that $k$ is algebraically closed here).

Remark: Actually, the image of the center of $H$ will be trivial unless $\text{char}(k)=2$.

To finish up with (3), observe that every elementary matrix in $\text{SL}_3(\mathbb{Z})$ is the center of a conjugate of $\text{H}(\mathbb{Z})$, thus the image of $\text{SL}_3(\mathbb{Z})$ consists of scalar matrices. But $\text{SL}_3(\mathbb{Z})$ is perfect, so this image is trivial.

(Edit: I edited a bit. Now the answer should be clearer, simpler and even correct. Thank you Andrei Smolensky for repeating correcting my embarrassing mistakes here.)

I am surprised that this old question was not fully answered yet. The answer is "No" and it is well known in some circles. In fact, a far more general statement holds:

1. Let $R$ and $S$ be rings (commutative with 1). Then any group homomorphism $\text{SL}_3(R)\to\text{SL}_2(S)$ factors via the non-trivial quotient $\text{SK}_1(3,R):=\text{SL}_3(R)/\text{EL}_3(R)$, where $\text{EL}_3(R)$ is the subgroup generated by elementary matrices in $\text{SL}_3(R)$.

Note that $\text{SL}_3(R)$ contains an epimorphic image of $\text{SL}_3(\mathbb{Z})$ (induced by the map $\mathbb{Z}\to R$). It is well known (and easy) that $\mathrm{EL}_3(\mathbb{Z})\simeq \mathrm{SL}_3(\mathbb{Z})$, thus the image of $\text{SL}_3(\mathbb{Z})$ is contained in $\mathrm{EL}_3(R)$, and in fact $\mathrm{EL}_3(R)$ is generated by $\text{SL}_3(\mathbb{Z})$ as a normal subgroup (as you can observe by playing with commutation relation of elementary matrices). Thus (1) is equivalent to:

2. Let $S$ be a ring (commutative with 1). Then any group homomorphism $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(S)$ is trivial.

We now fix a homomorphism as in statement (2) and assume its image is non-trivial. It is standard that there is a maximal ideal $m\lhd S$ and an integer $k$ such that the image of $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(S/m^k)$ is already non-trivial. Note that the the kernel of $\text{SL}_2(S/m^k)\to\text{SL}_2(S/m)$ is nilpotent and that $\text{SL}_3(\mathbb{Z})$ has no non-trivial nilpotent quotients (every elementary matrix is a commutator, thus it is perfect). It follows that $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(S/m)$ is non-trivial. We are left to prove the following statement:

3. Let $k$ be a field. Then any group homomorphism $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(k)$ is trivial.

(here I had before an argument I liked, but I had to replace it by a simpler one.)

Here is a nice exercise:

4. Let $k$ be a field. Then for any group homomorphism $\text{H}(\mathbb{Z})\to\text{SL}_2(k)$, where $\text{H}(\mathbb{Z})$ is the integral Heisenberg group, the image of the center (=commutator group) of $\text{H}(\mathbb{Z})$ consists of scalar matrices.

Hint: Assume the image of a generator of the center is not a scalar matrix and show that $\text{H}(\mathbb{Z})$ is in the Borel, in which every nilpotent group is abelian (you may assume that $k$ is algebraically closed here).

Remark: Actually, the image of the center of $\text{H}(\mathbb{Z})$ will be trivial unless $\text{char}(k)=2$.

To finish up with (3), observe that every elementary matrix in $\text{SL}_3(\mathbb{Z})$ is the center of a conjugate of $\text{H}(\mathbb{Z})$, thus the image of $\text{SL}_3(\mathbb{Z})$ consists of scalar matrices. But $\text{SL}_3(\mathbb{Z})$ is perfect, so this image is trivial.

I am surprised that this old question was not fully answered yet. The answer is "No" and it is well known in some circles. In fact, a far more general statement holds:

1. Let $R$ and $S$ be rings (commutative with 1). Then any group homomorphism $\text{SL}_3(R)\to\text{SL}_2(S)$ has a nilpotent image.

Note that $\text{SL}_3(R)$ contains $\text{SL}_3(\mathbb{Z})$ which is perfect, thus has no nilpotent (or solvable) quotients. This shows why (1) answers the asked question. I will use this fact more then once.

Recall that $\mathrm{EL}_3(R)<\text{SL}_3(R)$ (which denotes the subgroup generated by elementary matrices) is normal with a nilpotent cokernel (thank you, Andrei Smolensky, for correcting a mistake I made here earlier). Thus, to prove statement (1) it is enough to show that $\mathrm{EL}_3(R)$ is in the kernel of any homomorphism as above. Observe (by playing with commutation relations of elementary matrices) that the normal closure of the image of $\mathrm{EL}_3(\mathbb{Z})\to \mathrm{EL}_3(R)$ (induced by the map $\mathbb{Z}\to R$) is $\mathrm{EL}_3(R)$. Finally recall that $\mathrm{EL}_3(\mathbb{Z})\simeq \text{SL}_3(\mathbb{Z})$. Thus it is enough for us to prove the following statement:

Note that (2) is equivalent to (1) by the fact that $\text{SL}_3(\mathbb{Z})$ is perfect.

We now fix a homomorphism as in statement (2) and assume its image is non-trivial. It is standard that there is a maximal ideal $m\lhd S$ and an integer $k$ such that the image of $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(S/m^k)$ is already non-trivial. Note that the the kernel of $\text{SL}_2(S/m^k)\to\text{SL}_2(S/m)$ is nilpotent and that $\text{SL}_3(\mathbb{Z})$ has no non-trivial nilpotent quotients. It follows that $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(S/m)$ is non-trivial. We are left to prove the following statement:

Now there are many ways to proceed, and mine is not better than yours (please let me know your quick proof in a comment), but let me shoot with all the guns. There is a beatiful argument of Tits which I want to use. Since $\text{SL}_3(\mathbb{Z})$ is finitely generated, the matrix elements of its image generate a finitely generated domain in $k$, and this one could be embedded in a local field. Moreover, if the image of $\text{SL}_3(\mathbb{Z})$ in $\text{SL}_2(k)$ is infinite, this new embedding could be chosen such that the image of $\text{SL}_3(\mathbb{Z})$ is actually unbounded. Thus we get the following:

4. In (3) we can take $k$ to be a local field and assume that the image of the homomorphism is either unbounded or finite.

If the image of the homomorphism is unbounded we get a contradiction to Margulis' super-rigidity. So we may assume that the image is finite. In case $\text{char}(k)=0$ we may replace it with $\mathbb{C}$, thus assume that the image is in $\mathrm{SU}(2)$, which has no non-abelain nilpotent group, unlike any image of $\text{SL}_3(\mathbb{Z})$, and get a contradiction. So we get $k=\mathbb{F}_q((t))$ for some prime power $q$. The finite image of $\text{SL}_3(\mathbb{Z})$ is contained (up to conjugation) in the maximal compact $\text{SL}_2(\mathbb{F}_q[[t]])$. Arguing as we did after statement (2), with $S=\mathbb{F}_q[[t]]$ and $m=(t)$, we get:

5. In (3) we may assume $k$ is finite.

We now compose with the obvious map $\text{SL}_2(k)\to \text{PSL}_2(k)$. We should be a bit careful: recall the isomorphism $\text{SL}_3(\mathbb{F}_2)\simeq \text{PSL}_2(\mathbb{F}_7)$. (5) will follow from:

6. The only non-trivial homomorphism $\text{SL}_3(\mathbb{Z})\to \text{PSL}_2(\mathbb{F}_q)$ is for $q=7$.

Dickson classified all maximal subgroups of $\text{PSL}_2(\mathbb{F}_q)$ and all groups in the list which are non-solvable are of the form $\text{PSL}_2(\mathbb{F}_{q'})$ or $\text{PGL}_2(\mathbb{F}_{q'})$ for some prime power $q'<q$. By the fact that $\text{SL}_3(\mathbb{Z})$ is perfect we deduce that if its image is contained in $\text{PGL}_2(\mathbb{F}_{q'})$ then it is contained in its subgroup $\text{PSL}_2(\mathbb{F}_{q'})$. It follows that:

7. in (6) we can assume the homomorphism is onto.

Consider the image of the Heisenberg group. Since every non-abelian nilpotent subgroup of $\text{PSL}_2(\mathbb{F}_q)$ has an abelinization of order $2$, we conclude that the image of the elementary matrices of $\text{SL}_3(\mathbb{Z})$ are of order $2$. Therefore the homomorphism factors via a surjection $\text{SL}_3(\mathbb{F}_2)\to\text{PSL}_2(\mathbb{F}_q)$. The result follows by counting.

(Edit: I edited a bit. Now the answer should be clearer, simpler and even correct. Thank you Andrei Smolensky for repeating correcting my embarrassing mistakes here.)

I am surprised that this old question was not fully answered yet. The answer is "No" and it is well known in some circles. In fact, a far more general statement holds:

1. Let $R$ and $S$ be rings (commutative with 1). Then any group homomorphism $\text{SL}_3(R)\to\text{SL}_2(S)$ factors via the non-trivial quotient $\text{SK}_1(3,R):=\text{SL}_3(R)/\text{EL}_3(R)$, where $\text{EL}_3(R)$ is the subgroup generated by elementary matrices in $\text{SL}_3(R)$.

Note that $\text{SL}_3(R)$ contains an epimorphic image of $\text{SL}_3(\mathbb{Z})$ (induced by the map $\mathbb{Z}\to R$). It is well known (and easy) that $\mathrm{EL}_3(\mathbb{Z})\simeq \mathrm{SL}_3(\mathbb{Z})$, thus the image of $\text{SL}_3(\mathbb{Z})$ is contained in $\mathrm{EL}_3(R)$, and in fact $\mathrm{EL}_3(R)$ is generated by $\text{SL}_3(\mathbb{Z})$ as a normal subgroup (as you can observe by playing with commutation relation of elementary matrices). Thus (1) is equivalent to:

We now fix a homomorphism as in statement (2) and assume its image is non-trivial. It is standard that there is a maximal ideal $m\lhd S$ and an integer $k$ such that the image of $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(S/m^k)$ is already non-trivial. Note that the the kernel of $\text{SL}_2(S/m^k)\to\text{SL}_2(S/m)$ is nilpotent and that $\text{SL}_3(\mathbb{Z})$ has no non-trivial nilpotent quotients (every elementary matrix is a commutator, thus it is perfect). It follows that $\text{SL}_3(\mathbb{Z})\to\text{SL}_2(S/m)$ is non-trivial. We are left to prove the following statement:

(here I had before an argument I liked, but I had to replace it by a simpler one.)

Here is a nice exercise:

4. Let $k$ be a field. Then for any group homomorphism $H\to\text{SL}_2(k)$, where $H=\text{H}(\mathbb{Z})$ is the integral Heisenberg group, the image of the center (=commutator group) of $H$ consists of scalar matrices.

Hint: Assume the image of a generator of the center is not a scalar matrix and show that $H$ is in the Borel, in which every nilpotent group is abelian (you may assume that $k$ is algebraically closed here).

Remark: Actually, the image of the center of $H$ will be trivial unless $\text{char}(k)=2$.

To finish up with (3), observe that every elementary matrix in $\text{SL}_3(\mathbb{Z})$ is the center of a conjugate of $\text{H}(\mathbb{Z})$, thus the image of $\text{SL}_3(\mathbb{Z})$ consists of scalar matrices. But $\text{SL}_3(\mathbb{Z})$ is perfect, so this image is trivial.