Yes and no. A fairly precise statement of the pigeonhole principle would be:
If $A$ and $B$ are sets, and $A$ has more elements than $B$, and $f$ is a function $A\to B$, then $f$ is not injective.
Can it be proved? That depends. In particular, what does "$A$ has more elements than $B$" mean? In the usual development of set theory we use this phrase to mean the same as "$A$ has larger cardinality than $B$", which again means
There is an injective function $B\to A$, but there is no injective function $A\to B$.
So if we use that as our definition, the pigeonhole principle is not a matter of proof -- instead it is part of the definition of what it means for one set to be larger than the other.
Of course, once we define natural numbers, we might want to prove a "finite pigeonhole principle":
If $m$ and $n$ are naturals, and $m>n$, and $A$ has $m$ elements and $B$ has $n$ elements, and $f$ is a function $A\to B$, then $f$ is not injective.
Then all we have to do is to prove this from a definitions of "has $m$ elements" and "$m<n$".
The first of these is fairly easily done, because in the usual development, the natural number $m$ is represented by the set $\{0,1,2,\ldots,m-1\}$ and "has $m$ elements" means to be in bijective correspondence by precisely that set. So when we strip away those bijections, what we have to prove is that if $m>n$ and $f:m\to n$, then $f$ is not injective.
This would be trivial (and pointless) if we're using the cardinality definition of what $m>n$ means -- so even though that is the most common choice, let's assume that we want to define $m>n$ to mean "there is a natural $a$ such that $n+a+1=m$.
The actual content of the proof now is to show that these two competing definitions of $>$ agree! Of course we first need to define addition, but once we have done that, it is a fairly simple matter of induction.
First we prove that $|m+1|>|m|$ (as cardinals) for all $m$. The base case $m=0$ is easy. $0$ is the empty set, so there is no function for $1\to 0$ at all, so in the impossible case that we get one, we can safely claim it will be non-injective.
For the induction case, assume that $|m+1|>|m|$ and we need to show that $|m+1+1|>|m+1|$. Let $f: (m+1+1)\to(m+1)$ be given, and let $b=f(m+1)$
$$g:(m+1)\to m : x \mapsto\begin{cases}f(x) & \text{when }f(x)\ne m \\ b & \text{when }f(x)=m \end{cases}$$
Then by the induction hypothesis $g$ is not injective, so there is $p$ and $q$ with $g(p)=g(q)$. If $f(p)=f(q)$ then $f$ is not injective, and we're done. Otherwise $f(p)\ne f(q)$, but this can only be the case if one of them is $m$ and the other is $b$. But then either $f(p)$ or $f(q)$ equals $f(m+1)$, and $f$ is again not injective.
Now to complete the proof, we just have to handle the case where $a\ne 0$ in $|m+a+1|>|m+1|$. By this time we hopefully know that addition is commutative and associative, so $m+a+1=m+1+a$. And so if we have $f:(m+a+1)\to (m+1)$, then it is also $((m+1)+a)\to(m+1)$, and its restriction to $(m+1)$ is not injective. A restriction of an injective function would itself be injective, so $f$ is not injective. (Whew!).
(... except that we also ought to prove along the way that $p+a\subseteq p$ with the standard representation of the natural numbers; otherwise restricting the last $f$ to $(m+a)$ doesn't make sense).
For a third option, we could also have said that $m>n$ means that $n\in m$ for the set representation of the numbers. That would need a different proof from the induction above.
But all in all, these proofs are not very enlightening about the pigeonhole principle. Intuitively I would say that the pigeonhole principle is itself at least as obvious as it is that $n+a+1=m$ is a good definition of $m>n$. So what the proof actually proves could be argued to be just that the $n+a+1$ definition is reasonable. And this would also be the case for the $n\in m$ alternative.