I'm assuming the field to be algebrically closed
I know that there are other questions on this site that address the problem in the title, but the answers are just hints like: "you have to prove that the only regular functions are constants".
I'm following this advice, but I'm having some doubts.
Given a regular function $f$ I can restrict it to the affine space $U_0: x_0\neq 0$. Since the restriction of a regular function is regular $f|_{U_0}$ is a polynomial function of degree $d$ in the variables $(x_1/x_0,x_2/x_0)$:
$$f|_{U_0}=\frac{\sum_{|I|=d } a_Ix^I}{x_0^d}.$$
Since a regular function is entirely determined by its behaviour on an open subset, there is only a possible extension of $f|_{U_0}$.
Now I'd really like to say that this extension must necessarily be
$$f=\frac{\sum_{|I|=d } a_Ix^I}{x_0^d}.$$
So I can conclude that $d=0$, but I don't if this step is rigorous. And if it's rigorous, I don't know how to justify it properly.
Be careful, there is no reason for $f_{|U_0}$ to be homogeneous because $U_0$ is simply isomorphic to $\mathbb{A}^2$. The easiest way to look at $f_{|U_0}$ is to use the isomorphism $\varphi : (x_1,x_2) \mapsto [1:x_1:x_2]$ from $\mathbb{A}^2$ to $U_0$.
Now, you can write $f[1:x_1:x_2]$ as $\sum_{m \in S} a_mx_1^{m_1}x_2^{m_2}$ where $S \subset \mathbb{N}^2$ is finite. Thus, $f[x_0:x_1:x_2] = \sum_{m \in S} a_m\frac{x_1^{m_1}x_2^{m_2}}{x_0^{m_1 + m_2}}$ (but $m_1 + m_2$ has no reason to be constant when $m \in S$).
The idea now is that $\mathbb{P}^2\backslash\{[0:0:1]\}$ contains $U_0$ but also $U_1$. Therefore, you can also write $f[x_0:x_1:x_2] = \sum_{m \in T} b_m\frac{x_0^{m_1}x_2^{m_2}}{x_1^{m_1 + m_2}}$ when $x_1 \neq 0$.
You deduce that on $U_0 \cap U_1$, $$ \sum_{m \in S} a_m\frac{x_1^{m_1}x_2^{m_2}}{x_0^{m_1 + m_2}} = \sum_{m \in T} b_m\frac{x_0^{m_1}x_2^{m_2}}{x_1^{m_1 + m_2}}, $$ Since it is true on the whole non-empty open set $U_0 \cap U_1$, we obtain a formal equality between Laurent polynomials. If you look at it as a Laurent polynomial in $x_0$ with coefficients in $K[x_1,x_2,x_1^{-1}]$, you deduce from the valuation at $x_0$ that $\max(S) = 0$ thus $f$ is constant.