This is more a long comment, but it might be useful. A somehow overlooked fact is that

**Theorem.** For every function $u\in H^1(\mathbb R^2)$ and for every $\alpha\ge 0$, the integral 
$$
\int_{\mathbb R^2} (e^{\alpha u^2}-1) dx <+\infty
$$

That is, the integral is finite no matter what $\alpha$. Note however, that to ensure an uniform upper bound on the integral, $\alpha$ has to depend on (the norm of) $u$.

The proof of this fact easily follows the following three steps:

 1. By classical Moser Trudinger inequality, for every $\alpha >0$ there exists $R>0$ such that 
$$
\int_{\mathbb R^2} (e^{\alpha u^2}-1) dx \le 1
$$
for all $u \in B_R=\{u \in H^1: \|u\|_{H^1}\le R\}$.

2. For all $\alpha >0$ and for all $v\in C_c^\infty$
$$
\int_{\mathbb R^2} (e^{\alpha v^2}-1) dx <+\infty.
$$

3. Let $X$ be a normed space and let $F: X \to (-\infty,+\infty]$ be a convex function such that:

    (a). There exists $R>0$ such that $B_R\subset\{F\le 1\}$.

    (b). There exists a (norm) dense set $\mathcal D$ such that $F(v)<+\infty$ for all $v \in \mathcal D$.
   
   Then $F$ is finite everywhere.

Item 3 above should be well known, but I am not aware of any reference (and I would actually appreciate one if someone knows it). The proof goes as follow:

Let $u \in X$ and let $v \in \mathcal D$ and $w\in B_{2R}$ be such that 
$$
2u=v+w.
$$
Then 
$$
F(u)=F\Big(\frac{v+w}{2}\Big)\le \frac 1 2 F(v)+\frac 1 2 F(w)<+\infty.
$$