You could also apply Gaussian Elimination to get:
$$
\begin{pmatrix}
1 & c & c \\
0 & c^2 - c & c - c^2 \\
0 & 0 & -c + 3\\
\end{pmatrix}
$$
This matrix is singular if any element on the diagonal is zero, i.e, if:
$$ c^2 - c = 0 \text{ or } -c+3 = 0 $$
which is equivalent to computing the determinant..
In attempting to give you my complicated general formula to your question, a curious thing happened. All the zero terms made everything dissapear. This makes sense as $\mathbf{0}$ is obviously a solution. For a non-zero solution, $X$ and $Y$ must be similar matrices
$$XA=AY \Rightarrow A^{-1}XA = Y$$
Mainly for reference purposes, and to illustrate the subtle complexity of the equation, here is the $2 \times 2$ general form of the equation along with the solution (I used my own variables as I did not want to translate what I have, and the zeroes are the terms I mentioned):
\begin{array}{rcl}
\pmatrix{e & f \\ g & h}\pmatrix{q & r \\ s & t}=\pmatrix{q & r \\ s & t}\pmatrix{a & b \\ c & d} \\
\Rightarrow \pmatrix{q & r \\ s & t} &= & \pmatrix{\hat{q}\over {\Delta} & \hat{r}\over {\Delta} \\ \hat{s}\over {\Delta} & \hat{t}\over {\Delta}} \\
\end{array}
with
$$\Delta = a^2\left(d^2 -de -dh + eh -fg \right)
+ a\left(-2bcd + bce + bch -d^2e -d^2h + de^2 + 2deh + dh^2 -e^2h + efg -eh^2 + fgh \right) + bc\left( bc + de + dh -e^2 -2fg -h^2 \right)
+ d\left(deh -dfg -e^2h + efg -eh^2 + fgh \right)
+ e^2h^2 -2efgh + f^2g^2 $$
$$\hat{q} = 0 $$
$$\hat{r} = 0 $$
$$\hat{s} = 0 $$
$$\hat{t} =
0 $$
The formula for $\Delta$ here is a test for equal eigenvalues - it is zero if $X=\pmatrix{e & f \\ g & h}$ and $Y=\pmatrix{a & b \\ c & d}$ share at least one eigenvalue(not necessarily all eigenvalues).
Therefore, if $\Delta$ is non-zero, there is not a solution to your equation.
tl;dr: If you attempt to solve the equation by looking at the individual terms and getting simultaneous equations, you will be re-inventing a wheel called the Kronecker product. It is the tool you want to learn for this equation.
Best Answer
Hint: Compute $AB$ and $AC$, then solve a system to compute the entries of $C$.
One possible solution is letting $C =\begin{bmatrix} -1 & 1 \\ 1 & 3 \end{bmatrix}$ There are infinite solutions.
Edit: We have
$AB =\begin{bmatrix} 3 & 21 \\ 1 & 7 \end{bmatrix}$. If $C = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, then $AC=\begin{bmatrix} 3a+6c & 3b+6d \\ a+2c & b+2d \end{bmatrix}$.
So if $AB=AC$, it will follow that
Since 1. and 2. are equivalent, we'll work with 2. only. For the same reason we'll take 4. over 3.
Now we have that $a+2c=1$ and $b+2d=3$. Therefore $a=1-2c$ and $b=7-2d$.
This where I think you have trouble: you know $a$ and $b$, but you don't know $c$ and $d$. So what are $c,d$? As it happens $c,d$ can take any value you want! So just choose $c$ and $d$ as you wish, just as long as they satisfy the initial condition that all entries of $C$ are different from $0$. This means that $c$ can't be $0$ or $1\over 2$. Also $d$ can't be $0$ or $7\over 2$. I picked $c=1$ and $d=3$ to get the above mentioned solution.