Let $2^x+x^2+25=p^3$ where $p$ a prime number.$p^3>25>8$ so $p$ is odd. If $x<0$ then $2^x\not \in \mathbb{Z}$ so we must have $x\in \mathbb{N}$ $x=0\Rightarrow p^3=1+25=26$.So $x>0$ $2^x+x^2+25\equiv1\pmod2\Leftrightarrow x\equiv0\pmod 2$.Let $x=2x_1$ If $p=3$ then $27=25+2^x+x^2\geq 25+2+x^2>27$,contradiction So $p>3$.If $x\equiv 1,2\pmod 3$ then $p^3\equiv 2^{2x_1}+x^2+25\equiv 1+1+25\equiv 0\pmod 3$,contradiction So $3\mid x$.Let $x=3a$ ($2|a$) $p^3>2^{3a}\Rightarrow p\geq 2^a+1$. So $2^{3a}+9a^2+25\geq(2^a+1)^2=2^{3a}+3\cdot 4^a+3\cdot 2^a+1\Leftrightarrow 9a^2+25>3\cdot 4^a+3\cdot 2^a+1$ Lemma: $a\in \mathbb{N},a>3\Rightarrow 4^a>4a^2$. Proof: $4^4>4\cdot 4^2$.Let $4^k>4k^2$ for some $k>3$.Then $4^{k+1}=4\cdot4^k>4\cdot4k^2=4(2k)^2>4(k+1)^2$ So the lemma is true. If $a>3$ then $3\cdot 4^a+3\cdot2^a+1>12a^2+24+1>9a^2+25$,contradiction. So since $2|a$ we have that $a=2$. This is a solution because $x=3a=6\Rightarrow p^3=2^6+6^2+25=125=5^3$ and $5$ is prime So the only solution is $x=6$