Perhaps a Pell type equation Find all pairs of positive integers $(a,b)$ that satisfy $13^a+3=b^2$.
If $a$ is even then $3=(b-13^{a/2})(b+13^{a/2})$ which have no solutions. Now if the case $a=2k+1$ is odd then $b^2-13.(13^k)^{2}=3$ I cant proceed from here, please help. Any other methods for the latter case?

A solution using elliptic curves, or Mordell equations, as Will Jagy suggests in the comments.
There are three cases:
$1)$ $\ $ $a=3k$, $k\in\mathbb Z^+$. Then $\left(13^k\right)^3+3=b^2$ is a Mordell equation.
See http://oeis.org/search?q=mordell&language=english&go=Search,
in particular (this link is inside the link), e.g., http://oeis.org/A001014/a001014.txt
$b=\pm 2$, $13^k=1$, $k=0$, $a=0$. No solutions, because $k\in\mathbb Z^+$.
$2)$ $\ $ $a=3k+1$, $k\in\mathbb Z$, $k\ge 0$.
Then $\left(13^{k+1}\right)^3+507=(13b)^2$.
$13^{k+1}=13$, $13b=\pm 52$, $k=0$, $a=1$, $b=\pm 4$.
$3)$ $\ $ $a=3k+2$, $k\in\mathbb Z$, $k\ge 0$.
Then $3\cdot 13^4>10000$, so a different method is needed.
Then $13^3\equiv 1\pmod{61}$, $13^2\equiv 47\pmod{61}$
$\left(13^{3}\right)^k\cdot 13^2+3\equiv b^2\pmod{61}$
$50\equiv b^2\pmod{61}$
$2\equiv \left(b\cdot 5^{-1}\right)^2\pmod{61}$
because $\gcd(5,61)=1$.
A contradiction, because $2$ is not a quadratic residue mod $61$, because $61$ is a prime of the form $8k+5$. See Quadratic Reciprocity.
