Existence of positive integer k that are both squares Is there a positive integer k such that $4k+1$ and $9k+1$ are both squares?

Let's find the $k$ for which $4k+1 = a^2$ is a square. $4k+1$ is odd, so $a$ is odd. Then we have
$$k = \frac{a^2-1}{4} = \frac{a-1}{2}\cdot\frac{a+1}{2} = \frac{a-1}{2}\left(\frac{a-1}{2}+1\right)$$
the product of two consecutive numbers. For convenience, let's write $k =m\cdot (m+1)$. So can $9k+1$ be a square then?
$$9k+1 = 9m(m+1)+1 = 9m^2 + 9m + 1 = (3m+1)^2 + 3m = (3m+2)^2 - 3(m+1)$$
lies strictly between two consecutive squares then, unless $m = 0$ or $m+1 = 0$, which means $k = 0$, but that is excluded since $k$ was supposed to be positive.
