Here is the definition of a Sylow $p$-subgroup from Wikipedia:
For a prime number $p$, a Sylow $p$-subgroup (sometimes $p$-Sylow subgroup) of a group $G$ is a maximal $p$-subgroup of $G$, i.e., a subgroup of $G$ that is a $p$-group (so that the order of any group element is a power of $p$), and that is not a proper subgroup of any other $p$-subgroup of $G$.
The definition of a $p$-group is mentioned there in the brackets, but here is the definition as stated in its own article:
In mathematical group theory, given a prime number $p$, a $p$-group is a periodic group in which each element has a power of $p$ as its order: each element is of prime power order. That is, for each element $g$ of the group, there exists a nonnegative integer $n$ such that $g$ to the power $p^n$ is equal to the identity element.
My issue with these definitions concerns the trivial group. The group $\{\rm{id}\}$ has exactly one element of order $1$; since this is $p^0$ for any $p$, it follows that this group is a $p$-group for any prime $p$. In fact, this is the only group which is a $p$-group for more than one $p$, since if any element has an order greater than $1$, the prime that divides the order of this element must be the only $p$ for which $G$ is a $p$-group.
Given that the trivial group is a $p$-group, it also follows that nontrivial Sylow $p$-groups do not exist, because the trivial group is a subgroup of every group, so the only maximal $p$-group is the trivial group itself.
I am a bit bothered that no one seems to address this trivial case. Is the trivial group a $p$-group or not? Is it a Sylow $p$-group? Certainly the Sylow theorems are false under this definition, because there are no Sylow groups of size $p^n$ unless $n=0$.
I may be misunderstanding your point, but is it about what a "proper" subgroup is? The usual definition is that a subgroup $H\leq G$ is a proper subgroup if $H\neq G$, and in particular, the trivial subgroup is a proper subgroup of $G$ unless $G$ itself is the trivial group.