Induction.
According to the Peano Axioms in this article https://en.wikipedia.org/wiki/Peano_axioms, one axioms states that if $\varphi$ is a unary predicate such that $\varphi(0)$ is true and $$\forall n \in \mathbf{N}[\varphi(n) \implies \varphi(s(n))],$$ then $\forall n \in \mathbf{N}[P(n)]$ is true.
This does not allow $n$-ary predicates for $n \geq 2$, however I have seen that induction can be used for those as well. What is the formal reason for that?
Example.
Consider a formula of the form $\forall n (\forall m [(m \in \mathbf{N} \wedge n \in \mathbf{N}) \implies P(m,n)])$ (I hope this is a correct translation of "For all $n \in \mathbf{N}$ and for all $m \in \mathbf{N}$ the statement $P(m,n)$ is true"). One can prove this statement in the following way. Let $n \in \mathbf{N}$ and $m \in \mathbf{N}$. Then (...) and thus $P(n,m)$. Now, I have read that I can prove such a statement by induction the following way. Let $n \in \mathbf{N}$, that is, fix an element $n$. Prove $P(0,n)$ and for all $k \in \mathbf{N}$ with $P(k,n)$ show that $P(s(k),n)$ is true as well. Now, if I could apply induction to this, one could deduce $\forall m \in \mathbf{N} [P(m,n)] (*).$ Since $n$ was arbitrary, it holds that $\forall n (\forall m[m \in \mathbf{N} \wedge n \in \mathbf{N} \implies P(m,n)]).$
Questions.
The argument of my example, if correct, needs that induction can be applied to predicates of the form $P(m,n)$. $(1)$ Why can this be done/Why is this formally possible?
$(2)$ I have also read the following question Complete Induction on two variables, in which the answer defines $Q(x)=\forall y P(x,y)$. However, is $Q(x)$ then a valid unary predicate, that is, if $P(x,y)$ is a predicate, then $\forall y P(x,y)$ is a unary predicate?
$(3)$ I have also read the following question Double induction. Here the answer says that the formula $\forall m P(m,n)$ is not a sentence and can thus not be proven. However, in $(*)$ I have a similar expression. Did I do something wrong in my argument above?