Your professor is wrong, the solution is not unique. There are two (real-valued) solutions:
x1 = [2, 2, 0, 0]; y1 = [1, -1, 1, -1]
with DFTs
X1 = [4, 2-2i, 0, 2+2i]; Y1 = [0, 0, 4, 0]
(the product of which is obviously zero)
and
x2 = [2, -2, 0, 0]; y2 = [1, 1, 1, 1]
with DFTs
X2 = [0, 2+2i, 4, 2-2i]; Y2 = [4, 0, 0, 0]
(the product of which is also zero).
This is how you obtain these two solutions: first of all, note that the DFT of the circular convolution of the two sequences is equal to the product of the DFTs of the two sequences, where the individual DFTs are of length $4$. The product is zero for all indices $0\le k<4$ if for each index at least one of the two DFTs is zero. Since I assume that we're talking about real-valued sequences, we only need to consider the indices up to and including Nyquist (i.e. index $k=2$), because the last index $k=3$ is redundant due to the symmetry of the DFTs.
Since none of the two sequences can be identical to zero ($x[0]$ and $y[0]$ are both given and non-zero), only specific DFT bins can be zero, not all of them. For the length $4$ DFT of the zero-padded length $2$ sequence $x[n]$ it is easy to see that its DFT is given by
$$\begin{align}X[0]&=x[0]+x[1]\\
X[1]&=x[0]-jx[1]\\
X[2]&=x[0]-x[1]\\
X[3]&=x[0]+jx[1]\;\;(=X^*[1]\;)\end{align}\tag{1}$$
from which it is clear that with $x[0]\neq 0$, only $X[0]$ or $X[2]$ (but not both) can be zero. $X[1]$ can never be zero (for real-valued $x[1]$), from which it follows that $Y[1]$ must be zero. Since $Y[k]$ cannot be zero for all $k$, $X[0]$ or $X[2]$ must be zero. This leaves us with two possible choices for $x[1]$:
$$x[1]=x[0]\rightarrow X[2]=0\tag{a}$$
or
$$x[1]=-x[0]\rightarrow X[0]=0\tag{b}$$
In the first case, we need a DFT $Y[k]$ which is zero for all $k$ except for $k=2$, in the other case $Y[k]$ must be zero for all $k$ except for $k=0$. Note that $k=0$ corresponds to DC, and $k=2$ corresponds to Nyquist. The DFT of a sequence is zero everywhere except at DC if the sequence is constant, and it is zero everywhere except at Nyquist if the sequence is alternating. So for choice $(a)$ we need to choose $y[n]$ to alternate, and for choice $(b)$, $y[n]$ needs to be constant. These are the only two real-valued options.
EDIT: If you allow the sequences $x[n]$ and $y[n]$ to be complex-valued, there are two additional solutions. From $(1)$ it can be seen that for $x[0]\neq 0$, $X[k]$ can only be made zero for one index $k$. This means that $Y[k]$ must be zero for all indices except for that one index where $X[k]$ is zero. Therefore, $Y[k]$ must be a spectral line, i.e. its corresponding sequence $y[n]$ must be a complex exponential on the DFT ($N=4$) grid:
$$y[n]=e^{j\pi nl/2},\quad l=0,1,2,3\tag{2}$$
For $y[n]$ as given by $(2)$, $Y[k]$ is zero for $k=l$. The solutions $l=0$ (constant sequence), and $l=2$ alternating sequence are real-valued and were given above. The two other solutions are
y3 = [1, i, -1, -i]
($l=1$) and
y4 = [1, -i, -1, i]
($l=3$). The DFT of the corresponding sequences $x[n]$ must have a zero at $k=l$. From $(1)$ we get
x3 = [2, -2i, 0, 0]
and
x4 = [2, 2i, 0, 0]
The DFTs are
X3 = [2 - 2i, 0, 2 + 2i, 4]; Y3 = [0 4 0 0]
and
X4 = [2 + 2i, 4, 2 - 2i, 0]; Y4 = [0 0 0 4]
Both products are clearly zero.
In sum, there are four solutions to the problem, two of which are real-valued.
