Power of a prime number

Write $P = (x^2+1)(x+3).$ Notice that if $x=0$ this is a power of a prime. So now assume $x \ge 1.$ Suppose $P$ is a power of a prime.

In this case both $x^2+1$ and $x+3$ must be powers of the same prime. Since $x \ge 1,$ they are both divisible by that prime. But then so is $x^2+1-(x+3)(x-3) = 10.$ So the prime is either $2$ or $5.$

Suppose the prime is $2.$ We get $x = 2^k-3,$ and so $x^2+1 = 2^{2k} - 6 \cdot 2^k + 10$ is a power of $2.$ The first two terms are divisible by $4$ but the last term is not. So $x^2+1$ is not divisible by $4,$ and it's a power of $2,$ so it must equal $2.$ Hence $x=1,$ which is indeed a solution.

Suppose the prime is $5.$ We get $x=5^k-3,$ and so $x^2+1 = 5^{2k} - 6 \cdot 5^k + 10$ is a power of $5.$ If $k=1$ then $x=2,$ which is a solution. If $k \ge 2,$ the first two terms are divisible by $25$ but the last term is not. So $x^2+1$ is not divisible by $25,$ and it's a power of $5,$ so it must equal $5.$ This leads back to $x=2,$ which we have already discarded.

So the only solutions are $x=0,1,2,$ and the answer is $\fbox{3}.$