The question is as in the title. My thinking is that we should use the cohomology ring, since there is a similar nonexistence statement regarding just $\mathbb{C}P^2$.
The cohomology ring of $\mathbb{C}P^2\times\mathbb{C}P^2$ is, by the Kunneth formula,
$$
(\mathbb{Z}[x^2]/(x^{6})\otimes \mathbb{Z}[y^2]/(y^{6})
$$
(I use the squared variables to help me remember which degree the cohomology is in).
In particular, the top cohomology is generated by $x^{4}\otimes y^4=(x^2\otimes y^2)^2$, so if the induced map on cohomology sends $x^2\otimes y^2$ to say $H$, then $x^{4}\otimes y^4$ will be sent to $H^2$; by some case bashing, we can then show $H^2$ is never $-(x^4\otimes y^4)$. Is this enough to conclude the statement? I am used to thinking of the degree of a map in terms of the fundamental class in homology, and not cohomology. I think I am just missing a statement that can connect the two (for reference, answers should be geared towards a first year graduate course, say out of Hatcher). Thanks in advance.
Yes, it is enough to conclude.
Suppose that $f\colon X \longrightarrow Y$, $f_{*}\colon H_{n}(X)\longrightarrow H_{n}(Y)$, $f^{*}\colon H^{n}(Y)\longrightarrow H^{n}(X)$ . Then, if $H_{*}(X)$ and $H_{*}(Y)$ are free, $f_{*}$ and $f^{*}$ are dual.
Hence, if $f_{*}\colon H_{4}(\mathbb{C}P^2\times \mathbb{C}P^2 )\longrightarrow H_{4}(\mathbb{C}P^2\times \mathbb{C}P^2)$ sends $[\mathbb{C}P^2\times \mathbb{C}P^2]$ to $\text{deg }f [\mathbb{C}P^2\times \mathbb{C}P^2]$, $f^{*}$ sends the top cohomology $x^4 \otimes y^{4}$ to $\text{deg }f x^{4} \otimes y^{4}$
(I do not know if this answers your question.)