For all primes except $2$, $p^2\equiv 1$ mod $8\Rightarrow p^2+2453\equiv 0$ mod $8$. For all primes except $3$, $p^2\equiv 1$ mod $3\Rightarrow p^2+2453\equiv 0$ mod $3$. But now, suppose that some prime $p$ other than $2$ and $3$ divides $p^2+2453$. Then $2^3.3.p|p^2+2453$, and hence $p^2+2453$ has at least $16$ factors, so only $2$ and $3$ can be prime factors of $p^2+2453$. But the highest power of $3$ dividing $p^2+2453$ must be $2$, since if $2^33^3|p^2+2453$, we get at least $16$ factors. Similarly, the highest power of two dividing must be $6$, since $2^7.3|p^2+2453$ also gives at least $16$ factors. Combining these gives $p^2+2453\le2^63^2=576$ for $p\not\in \{2,3\}$, which is an obvious contradiction. It is easy to check that $p=3$ gives exactly $16$ factors, but $p=2$ gives $6$ factors, and hence $p=2$ is the unique valid solution.

This problem was used in a Thai competition two years ago. I think another solution develops from substituting $p = 6k+1, 6k-1$ and checking $p = 2,3$