Let's suppose that we have a random variable $Y$ and two probability measures $\mathbb{P}^0 \sim \mathbb{P}^1$ with Radon-Nikodym derivative $Z = \frac{d\mathbb{P}^1}{d\mathbb{P}^0}$. Let us also assume that $\mathbb{E}^{\mathbb{P}^0} Y^2 <\infty$.
Is there a condition to guarantee that $\mathbb{E}^{\mathbb{P}^1} Y^2 <\infty$ that is weaker than $\mathbb{P}^0 \left( Z<C \right)=1$ for some $0<C<\infty$?
I claim that if $Z$ is not $\Bbb P^1$-a.s. bounded by some constant $C$, then it always possible to construct a random variable $X$ on the underlying space $\Omega$ such that $\Bbb E^0|X|<\infty$ but $\Bbb E^1|X|=+\infty$.
Therefore, in general, the weakest possible condition on $Z$ to make sure all $\Bbb P^0$-integrable $X$ are also $\Bbb P^1$-integrable is the one you have stated, namely $\|Z\|_{L^{\infty}(\Bbb P^0)}<\infty$.
To prove this, suppose $\|Z\|_{L^{\infty}(\Bbb P^0)}=+\infty$. Let us define coefficients $a_k:= \frac{1}{k^2 P(Z > k)}.$ Then define the random variable $$X=\sum_{k=1}^{\infty} a_k \cdot 1_{\{Z>k\}}.$$Then note that $$\Bbb E^0|X| = \sum_k a_k P(Z>k) = \sum_k \frac{1}{k^2}<+\infty.$$However $$E^1|X| = E^0[XZ] =\sum_k a_k \Bbb E^0[Z \cdot 1_{\{Z>k\}}] \geq \sum_k ka_k P(Z>k) = \sum_k \frac{1}{k} = +\infty.$$
However, if you are willing to strengthen the condition on $Y$, then it is possible to weaken the condition on $Z$ (for instance if $Y \in L^{2q}(\Bbb P^0)$ for some $q >0$, then you only need $Z \in L^p(\Bbb P^0)$, where $p$ is the conjugate exponent to $q$).