By a theorem of Euler-Euclid, $n$ is an even perfect number iff $n=2^{p-1}(2^p-1)$, where $2^p-1$ is a (Mersenne) prime, thus so do $p$. Let $p=2$, then $n=6$ does not work. For $p=3$, we get $n=28$ works. I claim this is the only solution. Suppose $p\geqslant 5$. Now, $2^{p-1}(2^p-1)=a^3+b^3$. Note that, $a$ and $b$ has same parity, thus there are two cases to study. Suppose first that $a,b$ are both odd. Then, $2^{p-1}(2^p-1)=(a+b)(a^2-ab+b^2)$ implies, together with $a^2-ab+b^2$ is odd, that $a^2-ab+b^2=2^p-1$ and $a+b=2^{p-1}$. Hence, $a^2-ab+b^2 = 2(a+b)-1$. Using $a^2+b^2-ab\geqslant ab$, we have $(a-2)(b-2)\leqslant 3$. This is not too hard to investigate, and $a=3,b=1$ is the only solution; yielding $n=28=2^2(2^3-1)$. Let now $a,b$ both even, and set $a+b=2^\theta$ and $a^2-ab+b^2=2^{p-1-\theta}(2^p-1)$ (note that $2^p-1\nmid a+b$, as $a+b<a^2-ab+b^2$). Now, letting $b=2^\theta-a$, we arrive at $3a^2-3\cdot 2^\theta a+2^{2\theta} = a^2-ab+b^2 = (2^p-1)2^{p-1-\theta}$. Now, let $v_2(a)=i$. Since $a<2^\theta$, clearly $i<\theta$. Hence, $v_2(3a^2-3\cdot 2^\theta a+2^{2\theta})=2i=p-1-\theta$. Thus, $2^{\frac{p-1-\theta}{2}}\mid\mid a$. Now, using $(a+b)^2-(a^2-ab+b^2)=3ab$, we conclude that $3ab=2^{2\theta}-(2^p-1)2^{p-1-\theta}$. Consequently, $v_2(a+b)=p-1-\theta$. Thus, $2^{\frac{p-1-\theta}{2}}\mid b$. Let $a=2^{\frac{p-1-\theta}{2}} a_1$ and $b=2^{\frac{p-1-\theta}{2}}b_1$. With this move, we obtain $a_1^2-a_1b_1+b_1^2 = 2^p-1$. Moreover, $2^{\frac{3\theta-(p-1)}{2}}=a_1+b_1$. Thus, $a_1$ and $b_1$ have same parity. Now if both are even, so do $a_1^2-a_1b_1+b_1^2$, which clearly is not possible. Thus, both $a_1,b_1$ are odd. Now, if $a_1+b_1>6$, then $a_1+b_1\equiv 0\pmod{8}$. With this, we have $b_1\equiv -a_1\pmod{8}$, hence $a_1^2-a_1b_1+b_1^2\equiv 3a_1^2\equiv 3\pmod{8}$, whereas $2^p-1\equiv -1\pmod{8}$. This is a clear contradiction. Hence, $a_1+b_1\in\{2,4,6\}$ with $a_1,b_1$ being odd. If $a_1+b_1=2$, then $a_1^2-a_1b_1+b_1^2=1$, not a Mersenne prime. If $a_1+b_1=4$ with $a_1,b_1$ odd, then the only possibility is $a_1=3,b_1=1$ (or vice versa) in which case we have $a_1^2-a_1b_1+b_1^2=7$, that is $p=3$. Finally, for $a_1+b_1=6$, we have either $a_1=1,b_1=5$ (or vice versa), or $a_1=b_1=3$. In the former, $a_1^2-a_1b_1+b_1^2 = 21$, not a Mersenne prime. In the latter, $3\mid a_1^2-a_1b_1+b_1^2$, not possible. Thus, $n=28$ is indeed the only possibility.