You are right: assumptions of the (converse part) of the Wiener-Paley theorem can be substantially relaxed.
The usual formulation of this converse part says that if $F$ is an entire function of exponential type (say $\lambda>0$), and the restriction of $F$ on the real line is in $L^2$, then $F$ is a Fourier transform of an $L^2$ function with bounded support (bounded by $\lambda$).
In your problem, you know a priori that $f\in L^2$, and thus $F=\hat{f}\in L^2$, so the only condition needed is that $F$ is of exponential type. This means $$\log|F(z)|\leq\lambda|z|+o(z),\quad z\to\infty$$ Or, which is equivalent, $$\frac{1}{2\pi}\int_0^{2\pi}\log^+|F(re^{i\theta})|d\theta\leq (\lambda+o(1))r, \quad r\to\infty.$$ So, for example, if you know $|F(z)|\leq g(|x|)e^{\lambda|y|}$, where $g>0$, then it is sufficient to check $$\int_0^\pi\log^+ g(r\cos\theta)d\theta=o(r),\quad r\to\infty,$$ or even that this holds for some sequence $r_k\to\infty,$ to ensure that the support of $f$ is bounded by $\lambda$.
This can be further relaxed by using the Phragmen-Lindelof Principle. If you know that $f\in L^2$ on the real line, and also know that $\log|F(iy)|\leq \lambda|y|$, then it is enough to assume additionally that $$\log|F(z)|=O(|z|^{2-\epsilon}), \quad |z|=r_k\to\infty,$$ for some sequence $r_k$ to conclude that $F$ is in fact of exponential type $\lambda$.
For all these things I refer to the books of B. Y. Levin:
Lectures on entire functions, and
Distribution of zeros of entire functions.
Both are available on the web.