For $5b^2+4=n^2$, prove that $b=\frac{1}{5}(\lambda_2-\lambda_1)(\lambda_2^k-\lambda_1^k)$ where $\lambda_1, \lambda_2$ are roots of $x^2-3x+1=0$. If $b\in \mathbb{Z}_{>0}$ and $5b^2+4$ is a perfect square, then $b$ can be written in the form $b=\frac{1}{5}(\lambda_2-\lambda_1)(\lambda_2^k-\lambda_1^k)$, where $\lambda_1, \lambda_2$ are the two distinct solutions of the equation $x^2-3x+1=0$.
I have computed that $b=1,3,8, 21$ satisfies the condition and their $k$s are respectively $1,2,3,4$

COMMENT.-Your equation $5b^2+4=k^2$ is equivalent to $$\left(\dfrac k2\right)^2-5\left(\dfrac b2\right)^2=1$$ and the Pell-Fermat equation $$x^2-5y^2=1$$ has the infinite set of integer solutions where $n$ is natural integer.
$$x=\frac{(9+4\sqrt5)^n+(9-4\sqrt5)^n}{2}\\y=\frac{(9+4\sqrt5)^n-(9-4\sqrt5)^n}{2\sqrt5}$$ so you have integer solution for $k,b$
$$k=(9+4\sqrt5)^n+(9-4\sqrt5)^n\\b=\frac{(9+4\sqrt5)^n-(9-4\sqrt5)^n}{\sqrt5}$$ With  these solutions you can verify if it is true what you say.
