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Every positive powerful integer has a unique expression $a^2b^3$, with $a\ge1$ and $b\ge1$ squarefree. At a prime with even exponent put the whole exponent in $a^2$; at a prime with odd exponent, which is at least three, put three in $b^3$ and the remaining even exponent in $a^2$.

Fix squarefree $b,d$ in a representation $a^2b^3+c^2d^3=N$. Multiplying by $b^3$ gives

$$(b^3a)^2+(bd)^3c^2=b^3N=M.$$

Write $(bd)^3=Dq^2$ with $D$ positive squarefree. Then $\alpha=b^3a+qc\sqrt{-D}$ is an algebraic integer of norm $M$ in $K=\mathbb Q(\sqrt{-D})$. The map $(a,c)\mapsto\alpha$ is injective. In the ring of integers of $K$, unique factorization of nonzero ideals and $(\alpha)(\bar\alpha)=(M)$ show that $(\alpha)$ divides $(M)$. If $p^e\parallel M$, the numbers of ideal divisors of the $p$-part of $(M)$ in the split, inert and ramified cases are respectively $(e+1)^2$, $e+1$ and $2e+1$. Each is at most $(e+1)^2$. Thus $(M)$ has at most $\tau(M)^2$ ideal divisors, uniformly in $K$. Generators of the same principal ideal differ by a unit; an imaginary quadratic field has at most six units. There are therefore at most $6\tau(M)^2$ possible pairs $(a,c)$.

For clarity, the standard divisor estimate used here is uniform in its integer argument: for every $\eta>0$, $\tau(m)\ll_\eta m^\eta$. Indeed, for primes $p$ with $p^\eta\ge2$, one has $e+1\le2^e\le p^{\eta e}$. For each of the finitely many other primes, $(e+1)p^{-\eta e}$ is bounded over $e\ge0$; multiplying those bounds proves the assertion. Since $b^3\le N$ in any representation, $M\le N^2$. The fixed-$(b,d)$ count is consequently $N^{o(1)}$, with an implied bound independent of $b,d$ and $D$.

Put $a,b,c,d$ into half-open dyadic intervals with lower endpoints $A,B,C,E$, respectively. Any box containing a solution satisfies $A^2B^3\le N$ and $C^2E^3\le N$. Its number of solutions has each of the bounds

$$O(AB),\qquad O(CE),\qquad BE N^{o(1)}.$$

The first holds because $(a,b)$ determines the other summand and its canonical representation; the second is symmetric. The third sums the uniform norm bound over at most $O(BE)$ core pairs. Taking the geometric mean with weights $2/5,2/5,1/5$ bounds the minimum by

$$(AB)^{2/5}(CE)^{2/5}(BE)^{1/5}N^{o(1)} =(A^2B^3C^2E^3)^{1/5}N^{o(1)} \le N^{2/5+o(1)}.$$

There are $O((\log N)^4)$ nonempty dyadic parameter boxes. The subpower factor absorbs this count and is uniform across the boxes, proving the claim for every $N\ge2$ after enlarging the constant for small $N$.

This is a plus-sign argument. The real-quadratic difference equation in Section 2.5 of Tao's held version 2 is methodological context, not an application with a coefficient's sign silently changed. Its proof uses the same three-way dyadic interpolation; the norm argument above supplies the needed imaginary-quadratic interface.