Can $7n + 13$ ever equal a square? If not, why not? Can it be proved? And if it can be proved that it does equal a square (which I doubt), what is the smallest value for which this occurs?  

The answer by Prathyush Poduval is entirely correct, but the following perhaps goes a little deeper.
There can be at most $3$ non-zero quadratic residues $\pmod 7$. To see this, note that if $k$ is any non-negative integer and $n$ is any of the $6$ integers such that $0<n<7$, and $n^2 \equiv m \pmod 7$ then:
$$(7k+n)^2 \equiv 7(7k^2+2kn)+n^2 \equiv n^2 \equiv m \pmod 7$$
$$(7-n)^2 \equiv 7(7-2n)+n^2 \equiv n^2 \equiv m \pmod 7$$
Hence the number of non-zero quadratic residues is at most $6/2 = 3$.  To find all the non-zero quadratic residues it is therefore only necessary to consider:
$$1^2 \equiv 1 \pmod 7$$
$$2^2 \equiv 4 \pmod 7$$
$$3^2 \equiv 2 \pmod 7$$ 
