What is clear here is that $H_2$ has spectrum above $4$, and that $\lambda\in\sigma_{\textrm{ess}}(H_2)$ also, with $\lambda:=\max\sigma(H_2)=\|H_2\|>4$. What $\lambda$ is actually equal to seems a rather delicate question. Certainly $\lambda<20$, and I'm also not at all sure that $\lambda\ge 12$; that may already be too ambitious.

To discuss these claims, let me switch to the continuous counterpart of your problem, the operator $H=-\Delta-V(x,y)$ on $L^2(\mathbb R^2)$. Everything I'm going to say has a precise analog in your (discrete) setting, but the details become considerably less tedious in the continuous case. Even so, I will only sketch most of them.

We now assume that $V(x,y)=16$ on $|x|\le 1$, $y\ge 0$, and $V=0$ otherwise. We are interested in the part of the spectrum below zero. (We clearly have $[0,\infty)\subseteq\sigma (H)$.)

First of all, there actually is negative spectrum. This would not be clear in dimension $3$ or higher, but here it still works. Compare my answer here. We can find $\lambda=\min\sigma (H)$ as the infimum of the quadratic form $\lambda=\inf Q(f)$, $$ Q(f)=\int_{\mathbb R^2} \left(|\nabla f|^2-V|f|^2\right) , $$ taken over $f\in C_0^{\infty}$ (say) with $\|f\|_{L^2}=1$. Clearly, we can only hope to make $Q(f)$ negative if we concentrate $f$ where $V=16$, but then $Q(f)>-16$ because we can not avoid paying a (potentially steep) price in the first term. We can play around with this and get upper bounds on $\lambda$, but finding $\lambda$ exactly seems a formidable task.

Finally, there are $f\in L^2$ that make $\|(H-\lambda)f\|$ arbitrarily small. A carefully cut off version (compare the discussion in the comments to the linked answer) of $f$ will still give $Q(f)\simeq\lambda$, and then we obtain a Weyl sequence by also shifting and considering $g(x,y)=f(x,y-L)$, $L\gg 1$.

What is clear here is that $H_2$ has spectrum above $4$, and that $\lambda\in\sigma_{\textrm{ess}}(H_2)$ also, with $\lambda:=\max\sigma(H_2)=\|H_2\|>4$. What $\lambda$ is actually equal to seems a rather delicate question. Certainly $\lambda<20$, and I'm also not at all sure that $\lambda\ge 12$; that may already be too ambitious.

To discuss these claims, let me switch to the continuous counterpart of your problem, the operator $H=-\Delta-V(x,y)$ on $L^2(\mathbb R^2)$. Everything I'm going to say has a precise analog in your (discrete) setting, but the details become considerably less tedious in the continuous case. Even so, I will only sketch most of them.

We now assume that $V(x,y)=16$ on $|x|\le 1$, $y\ge 0$, and $V=0$ otherwise. We are interested in the part of the spectrum below zero. (We clearly have $[0,\infty)\subseteq\sigma (H)$.)

First of all, there actually is negative spectrum. This would not be clear in dimension $3$ or higher, but here it still works. Compare my answer here. We can find $\lambda=\min\sigma (H)$ as the infimum of the quadratic form $\lambda=\inf Q(f)$, $$ Q(f)=\int_{\mathbb R^2} \left(|\nabla f|^2-V|f|^2\right) , \tag{1} $$ taken over $f\in C_0^{\infty}$ (say) with $\|f\|_{L^2}=1$. Clearly, we can only hope to make $Q(f)$ small if we concentrate $f$ where $V=16$, but then $Q(f)>-16$ because we can not avoid paying a (potentially steep) price in the first term. A more chunky potential that is equal to $16$ on a large disk rather than a thin strip would be more effective at pulling down the bottom of the spectrum. In any event, we can try concrete test functions in (1) and get upper bounds on $\lambda$, but finding $\lambda$ exactly seems a formidable task.

Finally, there are $f\in L^2$ that make $\|(H-\lambda)f\|$ arbitrarily small. A carefully cut off version (compare the discussion in the comments to the linked answer) of $f$ will still keep $(H-\lambda)f$ small, and then we obtain a Weyl sequence by also shifting and considering $g(x,y)=f(x,y-L)$, $L\gg 1$.