It is easy to construct an infinite tree with any reasonable sequence as its degree sequence. (All ones, for example, would not be reasonable.)
For starters, here is a tree whose degree sequence, in your notation, is $(5^1,6^\infty)$. The vertex set is the set of all finite sequences $\langle x_1,x_2,\dots,x_n\rangle$ (including the null sequence) with $x_i\in\{0,1,2,3,4\}$, and with an edge joining $\langle x_1,x_2,\dots,x_n\rangle$ to $\langle x_1,x_2,\dots,x_n,x_{n+1}\rangle$. The null sequence has degree $5$, the rest have degree $6$.
By imposing the further conditions $x_1\ne0$ and $x_1\in\{1,2,3\}\implies x_2\in\{1,2,3\}$ we get a tree with degree sequence $(4^4,6^\infty)$; the vertices of degree $4$ are the null sequence and the sequences $\langle1\rangle$, $\langle2\rangle$, and $\langle3\rangle$.
The number of trees with degree sequence $(4^4,6^\infty)$ is $\aleph_0$; each such tree is determined up to isomorphism by the smallest (finite) subtree containing all the vertices of degree $4$, those vertices being marked.