Consider the adjacency matrix of the infinite d-regular tree, call it A. To find the spectrum we consider it as an operator in $L^2(V)$. It is stated that $A-\lambda I$ is always one-to-one. I do not see why this is true. This will tells us to find the spectrum we only need to find $\lambda$ such that $A-\lambda I$ is onto.
The only way I have convinced myself that this is true is by trying a bunch of examples, but I don't have a nice way of prove why this is true in generality.
Here's an idea. I don't know if it leads anywhere.
It's equivalent to show $\lambda I - A$ is always injective. What does an element of the kernel look like? Writing $E$ for the edge set, an element of the kernel is a list $(c_v)_{v \in V}$ of numbers $c_v$, one for each vertex $v$, such that $c_v$ is $\sum_{\{v,w\} \in E} c_w/\lambda$ for each $v$.
If an element of the kernel is nonzero, there is some nonzero $c_v$, and the surrounding $d$ vertices are weighted with numbers $c_w$ such that $\sum c_w/\lambda = c_v$. Looking at absolute values, it's not possible that all $d$ vertices have $|c_w| < |\lambda c_v|/d$, so some $w$ satisfies $|c_w| \geq |\lambda/d| |c_v|$. Repeat the same reasoning with the neighbors of $w$. If $|\lambda/d | > 1$, I think this shows you have a path $v = v_1, w = v_2, v_w,\ldots$ of vertices such that $|c_{v_j}| \to \infty$. I'm not sure what vector space you're using, but if it's the space of square-summable sequences $(c_v)_{v \in V}$, I'm pretty sure the absolute values of weights running off is bad news.
I don't know what you should do if $0 < |\lambda/d| \leq 1$, but if this idea actually meets your parameters, maybe an analogous idea will work for that case?
In any event, this is more along the lines of a tentative hint than an answer, but I hope it's slightly helpful.