Proving the existence of a proof without actually giving a proof
There are various ways to interpret the question. One interesting class of examples consists of "speed up" theorems. These generally involve two formal systems, $T_1$ and $T_2$, and family of statements which are provable in both $T_1$ and $T_2$, but for which the shortest formal proofs in $T_1$ are much longer than the shortest formal proofs in $T_2$.
One of the oldest such theorems is due to Gödel. He noticed that statements such as "This theorem cannot be proved in Peano Arithmetic in fewer than $10^{1000}$ symbols" are, in fact, provable in Peano Arithmetic.
Knowing this, we know that we could make a formal proof by cases that examines every Peano Arithmetic formal proof with fewer than $10^{1000}$ symbols and checks that none of them proves the statement. So we can prove indirectly that a formal proof of the statement in Peano Arithmetic exists.
But, because the statement is true, the shortest formal proof of the statement in Peano Arithmetic will in fact require more than $10^{1000}$ symbols. So nobody will be able to write out that formal proof completely. We can replace $10^{1000}$ with any number we wish, to obtain results whose shortest formal proof in Peano arithmetic must have at least that many symbols.
Similarly, if we prefer another formal system such as ZFC, we can consider statements such as "This theorem cannot be proved in ZFC in fewer than $10^{1000}$ symbols". In this way each sufficiently strong formal system will have some results which we know are formally provable, but for which the shortest formal proof in that system is too long to write down.
Henkin's completeness proof is an example of what you seek: It demonstrates that for a certain statement there is a proof, but does not establish what that proof is.
I'd like to expand Brian Tung's answer above, and add a further example.
In the following I'll be talking about first order (FO) sentences and structures. In a nutshell, a FO structure is a set $A$ with some operations and relations attached to it (e.g., a group, an ordered field, a graph), and a FO sentence is a statement you can write about $A$ with the restriction that $\forall x$ and $\exists x$ are only used for $x$ ranging over elements of $A$. Example from group theory are “$A$ is abelian”, “$A$ has exponent $2$”, but “$A$ has a finite subgroup” (“$\exists n\in\mathbb{N} : \dots$”) is not. An FO class is the class of all FO structures that satisfy a given set of FO sentences $\Phi$ (the class of all groups, all rings; less trivially, the class of $k$-vector spaces for a fixed $k$.)
Gödel's Completeness Theorem states that every FO sentence $\phi$ that holds in a FO class $\mathcal{M}$ has a formal proof from the axioms that define $\mathcal{M}$.
For the sake of clarity, let $\mathcal{M}$ the class of all rings. The important point to make is, no matter how you figured out that $\phi$ holds for every ring (using category theory, algebraic geometry or whatever), there must be some formal (finite!) proof that shows $\phi$ is a consequence of the axioms for $\mathcal{M}$.
Admittedly, having an FO proof of a statement would sound as something still too abstract (or useless) for an average mathematician. But we can push this example a little further.
It is a theorem by Jacobson that for each fixed $n\geq 2$, every ring that satisfies $\forall x: x^n=x$ is commutative. So if we take the axioms of rings and add $\forall x: x^{100}=x$, we obtain an FO class $\mathcal{M}'$ that satisfies $\forall x,y : xy = yx$. Actually, every axiom of $\mathcal{M}'$ is equational. Now Birkhoff's completeness theorem says the same as Gödel's but for equational proofs: Those that everyone here knows how to perform. That is, you start from the few axioms of $\mathcal{M}'$ and the only substantive rule to apply is
From $s=t$, conclude $p(s,y,z,\dots) = p(t,y,z,\dots)$,
where $p(x,y,z,\dots)$ is any expression built with the ring operations. So, despite Jacobson used other methods in his proof, for each $n$, there is purely elementary, simple-step, equational proof from the axioms of rings plus $\forall x: x^n=x$ of $\forall x,y : xy = yx$.
See this MO post for more on this example.