Is Toeplitz operator on the Bergman space bounded iff its symbol is bounded?

It is fairly well known that if $$T_\varphi$$ is a Toeplitz operator on the Hardy-Hilbert space, then $$\lVert T_\varphi \rVert = \lVert \varphi \rVert _{\infty}$$.

Now, if $$\varphi \in L^\infty (\mathbb D)$$ and $$T_{\varphi}$$ is a Toeplitz operator on the Bergman space then it is easy to see from the definition that $$\lVert T_\varphi \rVert \le \lVert {\varphi} \rVert_{\infty}$$. Is it also the case that bounded of the operator implies boundedness of the symbol? That is, if $$\varphi \in L^2 (\mathbb D)$$ is some symbol such that $$T_\varphi$$ is densely defined and bounded operator, do we have that $$\lVert \varphi \rVert_{\infty} \le M \lVert T_{\varphi} \rVert$$ for some $$M > 0$$?

I presume that people would have already studied these topics but I am unable to find so. References will be appreciated!

It is not neccesary in general that $$\varphi \in L^\infty(\mathbb{D})$$, but it is necessary and sufficient that in a certain sense $$\varphi$$ must be bounded on average in the hyperbolic sense''. The right condition for positive $$\varphi$$ is that there exists a constant $$C>0$$ such that
$$\begin{equation*} \frac{1}{(1-|w|)^2}\int_{D(z,(1-|z|)/2)}\varphi(w) dxdy \leq C,\,\, \text{for all}\,\, z \in \mathbb{D} \end{equation*}$$ where $$D(z,(1-|z|)/2)$$ is the Euclidean disc of center $$z$$ and radius $$(1-|z|)/2$$.
You can find this theorem in the Book of Zhu "Operator Theory in function spaces", Theorem 6.2.3. The idea for sufficiency is that $$T_\varphi(f) = P(\varphi f)$$, where $$P$$ is the Bergman projection. $$\begin{equation*} |\langle P(\varphi f), g \rangle |\leq \int_\mathbb{D} |\varphi(z) f(z) g(z) | dxdy \le \Big( \int_\mathbb{D} |f(z)|^2 \varphi(z) dxdy \Big)^{\frac{1}{2}} \Big( \int |g(z)|^2\varphi(z) dxdy \Big)^{\frac{1}{2}}. \end{equation*}$$ So if the measure $$\varphi(z)dxdy$$ is a Carleson measure for the Bergman space we have that $$\begin{equation*} |\langle T_\varphi(f), g \rangle | \leq C_\varphi \Vert f\Vert \Vert g\Vert. \end{equation*}$$ In particular the function $$\varphi(z) = \log\frac{1}{|z|}$$ gives a bounded Toeplitz operator but it is not a bounded function.