The Fibonacci numbers ($F_0=0$, $F_1=1$, $F_{n}=F_{n-1}+F_{n-2}$) have the identity $$F_{2k+1}=F_k^2 + F_{k+1}^2.$$ In particular, if $n$ is odd, then $F_n$ is a sum of two squares. Are there infinitely many even $n$ for which $F_n$ is a sum of two squares?

Some comments (building from my notes and also from the comments and answers below)

The set of even $n$ (nonnegative, naturally) with $F_{2n}$ the sum of two squares begins $$\{ 0,2,6,12,14,26,38,62,74,86,98,122,134,146,158,182,222,254,326,338,366,398,446,\cdots\},$$ using the tables at Mersennus.net. There are five entries in this list that are not 2 mod 12: three small numbers (0, 6, 12) that can be forgiven their impertinence, but also 222 and 366 (both are 6 mod 12, and also 78 mod 144).

The Fibonacci numbers are periodic modulo $m$ (for any $m>1$). Considering the sequence modulo 4, for example, it repeats 0, 1, 1, 2, 3, 1. Since the sum of two squares is never 3 mod 4, we learn that $F_{6n+4}$ is never the sum of two squares. Varying the modulus allows us to eliminate many other congruence classes. There are some numbers, for example 78, that are not the sum of two squares but do not seem to be eliminatable in this manner.