We want $8n + 1 = k^2$ for some integer $k$.

Then $8n = k^2 - 1 = (k - 1)(k + 1)$.

So $n = \dfrac{(k - 1)(k + 1)}{8}$.

We require $n$ to be an integer with $1 \leq n \leq 800$.

Let us analyze when $n$ is an integer.

Note that $k$ must be odd, since $k^2 \equiv 1 \pmod{8}$ only when $k$ is odd.

Let $k = 2m + 1$ for integer $m \geq 1$.

Then:

$k - 1 = 2m$, $k + 1 = 2m + 2$

So $(k - 1)(k + 1) = 2m \cdot (2m + 2) = 4m(m + 1)$

Therefore,

$n = \dfrac{4m(m + 1)}{8} = \dfrac{m(m + 1)}{2}$

We require $1 \leq n \leq 800$.

So $1 \leq \dfrac{m(m + 1)}{2} \leq 800$

Multiply both sides by $2$:

$2 \leq m(m + 1) \leq 1600$

Now, $m(m + 1)$ increases rapidly. Let's find the largest $m$ such that $m(m + 1) \leq 1600$.

Solve $m^2 + m - 1600 \leq 0$

The positive root is $m = \dfrac{-1 + \sqrt{1 + 6400}}{2} = \dfrac{-1 + 80.00625...}{2} \approx 39.5$

So $m$ can be at most $39$.

Now, $m$ must be a positive integer such that $m(m + 1) \geq 2$.

For $m = 1$, $1 \cdot 2 = 2$.

So $m$ runs from $1$ to $39$ inclusive.

Thus, the number of such $n$ is $39$.

Answer: $39$